4.2 Ordinary & Partial Differentiation

Detailed Theory: Ordinary & Partial Differentiation

1. Introduction to Derivatives

1.1 What is a Derivative?

The derivative measures how a function changes as its input changes. It represents the instantaneous rate of change or the slope of the tangent line to the curve at a point.

1.2 Definition of Derivative

The derivative of a function f(x)f(x) at point x=ax = a is:

f(a)=limh0f(a+h)f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

Alternative notation: f(x)f'(x), dfdx\frac{df}{dx}, ddxf(x)\frac{d}{dx}f(x)

1.3 Geometric Interpretation

  • f(a)f'(a) = slope of tangent line to y=f(x)y = f(x) at x=ax = a

  • If f(a)>0f'(a) > 0: function increasing at x=ax = a

  • If f(a)<0f'(a) < 0: function decreasing at x=ax = a

  • If f(a)=0f'(a) = 0: possible local maximum/minimum or inflection point

1.4 Example: Finding Derivative from Definition

Find derivative of f(x)=x2f(x) = x^2 using definition.

Solution:

f(x)=limh0(x+h)2x2hf'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}
=limh0x2+2xh+h2x2h= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}
=limh02xh+h2h= \lim_{h \to 0} \frac{2xh + h^2}{h}
=limh0(2x+h)=2x= \lim_{h \to 0} (2x + h) = 2x

So f(x)=2xf'(x) = 2x


2. Basic Differentiation Rules

2.1 Power Rule

For any real number nn:

ddx(xn)=nxn1\frac{d}{dx}(x^n) = nx^{n-1}

Examples:

  • ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2

  • ddx(x)=ddx(x1/2)=12x1/2=12x\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}

  • ddx(1x)=ddx(x1)=x2=1x2\frac{d}{dx}\left(\frac{1}{x}\right) = \frac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2}

2.2 Constant Rule

For constant cc:

ddx(c)=0\frac{d}{dx}(c) = 0

2.3 Constant Multiple Rule

For constant cc:

ddx[cf(x)]=cf(x)\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)

Example: ddx(5x3)=53x2=15x2\frac{d}{dx}(5x^3) = 5 \cdot 3x^2 = 15x^2

2.4 Sum/Difference Rule

ddx[f(x)±g(x)]=f(x)±g(x)\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)

Example: ddx(x2+3x5)=2x+3\frac{d}{dx}(x^2 + 3x - 5) = 2x + 3

2.5 Product Rule

ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)

Example: Find ddx(x2sinx)\frac{d}{dx}(x^2 \sin x)

ddx(x2sinx)=(2x)(sinx)+(x2)(cosx)=2xsinx+x2cosx\frac{d}{dx}(x^2 \sin x) = (2x)(\sin x) + (x^2)(\cos x) = 2x \sin x + x^2 \cos x

2.6 Quotient Rule

ddx[f(x)g(x)]=f(x)g(x)f(x)g(x)[g(x)]2(g(x)0)\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \quad (g(x) \neq 0)

Example: Find ddx(xx2+1)\frac{d}{dx}\left(\frac{x}{x^2 + 1}\right)

ddx(xx2+1)=(1)(x2+1)(x)(2x)(x2+1)2\frac{d}{dx}\left(\frac{x}{x^2 + 1}\right) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2}
=x2+12x2(x2+1)2=1x2(x2+1)2= \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}

2.7 Chain Rule

For composite function f(g(x))f(g(x)):

ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

Alternative notation: If y=f(u)y = f(u) and u=g(x)u = g(x), then:

dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Example: Find ddx[sin(x2)]\frac{d}{dx}[\sin(x^2)]

Let u=x2u = x^2, then y=sinuy = \sin u

dydu=cosu=cos(x2)\frac{dy}{du} = \cos u = \cos(x^2)
dudx=2x\frac{du}{dx} = 2x

So dydx=cos(x2)2x=2xcos(x2)\frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x \cos(x^2)


3. Derivatives of Elementary Functions

3.1 Trigonometric Functions

  1. ddx(sinx)=cosx\frac{d}{dx}(\sin x) = \cos x

  2. ddx(cosx)=sinx\frac{d}{dx}(\cos x) = -\sin x

  3. ddx(tanx)=sec2x\frac{d}{dx}(\tan x) = \sec^2 x

  4. ddx(cotx)=csc2x\frac{d}{dx}(\cot x) = -\csc^2 x

  5. ddx(secx)=secxtanx\frac{d}{dx}(\sec x) = \sec x \tan x

  6. ddx(cscx)=cscxcotx\frac{d}{dx}(\csc x) = -\csc x \cot x

3.2 Exponential and Logarithmic Functions

  1. ddx(ex)=ex\frac{d}{dx}(e^x) = e^x

  2. ddx(ax)=axlna(a>0,a1)\frac{d}{dx}(a^x) = a^x \ln a \quad (a > 0, a \neq 1)

  3. ddx(lnx)=1x(x>0)\frac{d}{dx}(\ln x) = \frac{1}{x} \quad (x > 0)

  4. ddx(logax)=1xlna(a>0,a1)\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a} \quad (a > 0, a \neq 1)

3.3 Inverse Trigonometric Functions

  1. ddx(sin1x)=11x2(1<x<1)\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \quad (-1 < x < 1)

  2. ddx(cos1x)=11x2(1<x<1)\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1 - x^2}} \quad (-1 < x < 1)

  3. ddx(tan1x)=11+x2\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2}

  4. ddx(cot1x)=11+x2\frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1 + x^2}

  5. ddx(sec1x)=1xx21(x>1)\frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2 - 1}} \quad (|x| > 1)

  6. ddx(csc1x)=1xx21(x>1)\frac{d}{dx}(\csc^{-1} x) = -\frac{1}{|x|\sqrt{x^2 - 1}} \quad (|x| > 1)

3.4 Hyperbolic Functions

  1. ddx(sinhx)=coshx\frac{d}{dx}(\sinh x) = \cosh x

  2. ddx(coshx)=sinhx\frac{d}{dx}(\cosh x) = \sinh x

  3. ddx(tanhx)=sech2x\frac{d}{dx}(\tanh x) = \text{sech}^2 x


4. Higher Order Derivatives

4.1 Definition

The second derivative is the derivative of the first derivative:

f(x)=ddx[f(x)]=d2fdx2f''(x) = \frac{d}{dx}[f'(x)] = \frac{d^2f}{dx^2}

The n-th derivative:

f(n)(x)=dnfdxnf^{(n)}(x) = \frac{d^n f}{dx^n}

4.2 Notation

  • First derivative: f(x)f'(x), f(1)(x)f^{(1)}(x), dfdx\frac{df}{dx}

  • Second derivative: f(x)f''(x), f(2)(x)f^{(2)}(x), d2fdx2\frac{d^2f}{dx^2}

  • n-th derivative: f(n)(x)f^{(n)}(x), dnfdxn\frac{d^n f}{dx^n}

4.3 Physical Interpretation

  • Position → Velocity → Acceleration:

    • If s(t)s(t) = position, then v(t)=s(t)v(t) = s'(t) = velocity

    • a(t)=v(t)=s(t)a(t) = v'(t) = s''(t) = acceleration

  • Curvature: Second derivative gives information about concavity

    • f(x)>0f''(x) > 0: concave up

    • f(x)<0f''(x) < 0: concave down

    • f(x)=0f''(x) = 0: possible inflection point

4.4 Example

Find first three derivatives of f(x)=x33x2+2xf(x) = x^3 - 3x^2 + 2x

f(x)=3x26x+2f'(x) = 3x^2 - 6x + 2
f(x)=6x6f''(x) = 6x - 6
f(x)=6f'''(x) = 6

5. Implicit Differentiation

5.1 What is Implicit Differentiation?

Used when yy is defined implicitly as a function of xx by an equation, not explicitly as y=f(x)y = f(x).

5.2 Method

  1. Differentiate both sides of equation with respect to xx

  2. Treat yy as function of xx (use chain rule for yy terms)

  3. Solve for dydx\frac{dy}{dx}

5.3 Examples

Example 1: Find dydx\frac{dy}{dx} for x2+y2=25x^2 + y^2 = 25

Differentiate both sides:

ddx(x2)+ddx(y2)=ddx(25)\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)
2x+2ydydx=02x + 2y \frac{dy}{dx} = 0

Solve for dydx\frac{dy}{dx}:

2ydydx=2x2y \frac{dy}{dx} = -2x
dydx=xy\frac{dy}{dx} = -\frac{x}{y}

Example 2: Find dydx\frac{dy}{dx} for x3+y3=6xyx^3 + y^3 = 6xy

Differentiate:

3x2+3y2dydx=6y+6xdydx3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}

Rearrange:

3y2dydx6xdydx=6y3x23y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2
(3y26x)dydx=6y3x2(3y^2 - 6x) \frac{dy}{dx} = 6y - 3x^2
dydx=6y3x23y26x=2yx2y22x\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}

6. Logarithmic Differentiation

6.1 When to Use

Useful for:

  1. Functions of form y=[f(x)]g(x)y = [f(x)]^{g(x)}

  2. Products/quotients with many factors

  3. Functions with variables in both base and exponent

6.2 Method

  1. Take natural logarithm of both sides: lny=ln[f(x)]\ln y = \ln[f(x)]

  2. Simplify using logarithm properties

  3. Differentiate implicitly with respect to xx

  4. Solve for dydx\frac{dy}{dx}

6.3 Examples

Example 1: Find dydx\frac{dy}{dx} for y=xxy = x^x

Take natural log: lny=ln(xx)=xlnx\ln y = \ln(x^x) = x \ln x

Differentiate implicitly:

1ydydx=lnx+x1x=lnx+1\frac{1}{y} \frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1
dydx=y(lnx+1)=xx(lnx+1)\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)

Example 2: Find dydx\frac{dy}{dx} for y=(x+1)2x1(x+3)3y = \frac{(x+1)^2 \sqrt{x-1}}{(x+3)^3}

Take natural log:

lny=2ln(x+1)+12ln(x1)3ln(x+3)\ln y = 2\ln(x+1) + \frac{1}{2}\ln(x-1) - 3\ln(x+3)

Differentiate:

1ydydx=2x+1+12(x1)3x+3\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+3}
dydx=y[2x+1+12(x1)3x+3]\frac{dy}{dx} = y\left[\frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+3}\right]
=(x+1)2x1(x+3)3[2x+1+12(x1)3x+3]= \frac{(x+1)^2 \sqrt{x-1}}{(x+3)^3} \left[\frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+3}\right]

7. Applications of Derivatives

7.1 Tangent and Normal Lines

  • Tangent line at x=ax = a: yf(a)=f(a)(xa)y - f(a) = f'(a)(x - a)

  • Normal line at x=ax = a: yf(a)=1f(a)(xa)y - f(a) = -\frac{1}{f'(a)}(x - a) (if f(a)0f'(a) \neq 0)

Example: Find tangent and normal to y=x2y = x^2 at (2,4)(2, 4)

f(x)=2xf'(x) = 2x, so f(2)=4f'(2) = 4

Tangent: y4=4(x2)y - 4 = 4(x - 2) or y=4x4y = 4x - 4

Normal: y4=14(x2)y - 4 = -\frac{1}{4}(x - 2) or y=14x+92y = -\frac{1}{4}x + \frac{9}{2}

7.2 Rates of Change

If y=f(x)y = f(x), then dydx\frac{dy}{dx} = instantaneous rate of change of yy with respect to xx.

Example: Radius of circle increasing at 3 cm/s. How fast is area changing when radius = 10 cm?

A=πr2A = \pi r^2, drdt=3\frac{dr}{dt} = 3

dAdt=dAdrdrdt=2πr3=6πr\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2\pi r \cdot 3 = 6\pi r

When r=10r = 10: dAdt=6π(10)=60π cm2/s\frac{dA}{dt} = 6\pi(10) = 60\pi \text{ cm}^2/\text{s}

Problems where two or more quantities are related and changing over time.

Strategy:

  1. Identify given rates and wanted rate

  2. Find equation relating variables

  3. Differentiate with respect to time

  4. Substitute known values and solve

Example: Ladder 10 ft long slides down wall. Bottom moves away at 1 ft/s. How fast is top sliding down when bottom is 6 ft from wall?

Let xx = distance from wall, yy = height on wall

Given: x2+y2=100x^2 + y^2 = 100, dxdt=1\frac{dx}{dt} = 1, find dydt\frac{dy}{dt} when x=6x = 6

Differentiate: 2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

When x=6x = 6, y=10036=64=8y = \sqrt{100 - 36} = \sqrt{64} = 8

2(6)(1)+2(8)dydt=02(6)(1) + 2(8)\frac{dy}{dt} = 0
12+16dydt=012 + 16\frac{dy}{dt} = 0
dydt=1216=34 ft/s\frac{dy}{dt} = -\frac{12}{16} = -\frac{3}{4} \text{ ft/s}

Negative means top is sliding down.


8. Partial Differentiation

8.1 Functions of Several Variables

A function of two variables: z=f(x,y)z = f(x, y)

Example: f(x,y)=x2+2xy+y2f(x, y) = x^2 + 2xy + y^2

8.2 Partial Derivatives

  • Partial derivative with respect to xx: Treat yy as constant Notation: fxf_x, fx\frac{\partial f}{\partial x}, xf(x,y)\frac{\partial}{\partial x}f(x, y)

  • Partial derivative with respect to yy: Treat xx as constant Notation: fyf_y, fy\frac{\partial f}{\partial y}, yf(x,y)\frac{\partial}{\partial y}f(x, y)

8.3 Computing Partial Derivatives

Example: For f(x,y)=x2y+sin(xy)f(x, y) = x^2y + \sin(xy)

fx=fx=2xy+ycos(xy)f_x = \frac{\partial f}{\partial x} = 2xy + y\cos(xy)
fy=fy=x2+xcos(xy)f_y = \frac{\partial f}{\partial y} = x^2 + x\cos(xy)

8.4 Higher Order Partial Derivatives

For f(x,y)f(x, y):

  1. Second partials:

    • fxx=x(fx)=2fx2f_{xx} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial^2 f}{\partial x^2}

    • fyy=y(fy)=2fy2f_{yy} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial^2 f}{\partial y^2}

    • fxy=y(fx)=2fyxf_{xy} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial^2 f}{\partial y \partial x}

    • fyx=x(fy)=2fxyf_{yx} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial^2 f}{\partial x \partial y}

  2. Clairaut's Theorem: If fxyf_{xy} and fyxf_{yx} are continuous, then fxy=fyxf_{xy} = f_{yx}

8.5 Example: Higher Order Partials

For f(x,y)=ex2+yf(x, y) = e^{x^2 + y}

fx=2xex2+yf_x = 2xe^{x^2 + y}
fy=ex2+yf_y = e^{x^2 + y}
fxx=x(2xex2+y)=2ex2+y+4x2ex2+y=2ex2+y(1+2x2)f_{xx} = \frac{\partial}{\partial x}(2xe^{x^2 + y}) = 2e^{x^2 + y} + 4x^2e^{x^2 + y} = 2e^{x^2 + y}(1 + 2x^2)
fyy=y(ex2+y)=ex2+yf_{yy} = \frac{\partial}{\partial y}(e^{x^2 + y}) = e^{x^2 + y}
fxy=y(2xex2+y)=2xex2+yf_{xy} = \frac{\partial}{\partial y}(2xe^{x^2 + y}) = 2xe^{x^2 + y}
fyx=x(ex2+y)=2xex2+yf_{yx} = \frac{\partial}{\partial x}(e^{x^2 + y}) = 2xe^{x^2 + y}

Note: fxy=fyxf_{xy} = f_{yx}


9. Total Differential and Chain Rule for Partial Derivatives

9.1 Total Differential

For z=f(x,y)z = f(x, y), the total differential is:

dz=zxdx+zydydz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy

Interpretation: Approximate change in zz when xx changes by dxdx and yy changes by dydy.

9.2 Chain Rule for Partial Derivatives

Case 1: z=f(x,y)z = f(x, y), x=g(t)x = g(t), y=h(t)y = h(t)

dzdt=zxdxdt+zydydt\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}

Case 2: z=f(x,y)z = f(x, y), x=g(s,t)x = g(s, t), y=h(s,t)y = h(s, t)

zs=zxxs+zyys\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}
zt=zxxt+zyyt\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}

9.3 Example of Chain Rule

Let z=x2yz = x^2y, x=s+tx = s + t, y=sty = st

Find zs\frac{\partial z}{\partial s} and zt\frac{\partial z}{\partial t}

First: zx=2xy\frac{\partial z}{\partial x} = 2xy, zy=x2\frac{\partial z}{\partial y} = x^2

Also: xs=1\frac{\partial x}{\partial s} = 1, xt=1\frac{\partial x}{\partial t} = 1, ys=t\frac{\partial y}{\partial s} = t, yt=s\frac{\partial y}{\partial t} = s

Now:

zs=(2xy)(1)+(x2)(t)=2xy+tx2\frac{\partial z}{\partial s} = (2xy)(1) + (x^2)(t) = 2xy + tx^2

Substitute x=s+tx = s+t, y=sty = st:

=2(s+t)(st)+t(s+t)2=2s2t+2st2+t(s2+2st+t2)= 2(s+t)(st) + t(s+t)^2 = 2s^2t + 2st^2 + t(s^2 + 2st + t^2)
=2s2t+2st2+ts2+2st2+t3=3s2t+4st2+t3= 2s^2t + 2st^2 + ts^2 + 2st^2 + t^3 = 3s^2t + 4st^2 + t^3

Similarly:

zt=(2xy)(1)+(x2)(s)=2xy+sx2\frac{\partial z}{\partial t} = (2xy)(1) + (x^2)(s) = 2xy + sx^2

Substitute: =2(s+t)(st)+s(s+t)2=2s2t+2st2+s3+2s2t+st2= 2(s+t)(st) + s(s+t)^2 = 2s^2t + 2st^2 + s^3 + 2s^2t + st^2

=s3+4s2t+3st2= s^3 + 4s^2t + 3st^2

10. Gradient and Directional Derivatives

10.1 Gradient Vector

For f(x,y)f(x, y), the gradient is:

f=fx,fy\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle

For f(x,y,z)f(x, y, z):

f=fx,fy,fz\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle

10.2 Properties of Gradient

  1. Points in direction of maximum increase of ff

  2. Magnitude = maximum rate of increase

  3. Perpendicular to level curves/surfaces

10.3 Directional Derivative

Rate of change of ff in direction of unit vector u=a,b\vec{u} = \langle a, b \rangle:

Duf=fu=fxa+fybD_{\vec{u}} f = \nabla f \cdot \vec{u} = f_x a + f_y b

Example: Find directional derivative of f(x,y)=x2+y2f(x, y) = x^2 + y^2 at (1,2)(1, 2) in direction of v=3,4\vec{v} = \langle 3, 4 \rangle

First, make v\vec{v} unit vector: v=9+16=5\|\vec{v}\| = \sqrt{9 + 16} = 5

u=35,45\vec{u} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle

Gradient: f=2x,2y\nabla f = \langle 2x, 2y \rangle

At (1,2)(1, 2): f(1,2)=2,4\nabla f(1, 2) = \langle 2, 4 \rangle

Directional derivative:

Duf=2,435,45=65+165=225D_{\vec{u}} f = \langle 2, 4 \rangle \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle = \frac{6}{5} + \frac{16}{5} = \frac{22}{5}

11. Taylor Series and Maclaurin Series

11.1 Taylor Series for Single Variable

For f(x)f(x) about x=ax = a:

f(x)=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3+f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots
=n=0f(n)(a)n!(xa)n= \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

11.2 Maclaurin Series

Taylor series about a=0a = 0:

f(x)=f(0)+f(0)x+f(0)2!x2+f(0)3!x3+f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots

11.3 Important Maclaurin Series

  1. ex=1+x+x22!+x33!+=n=0xnn!e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}

  2. sinx=xx33!+x55!x77!+=n=0(1)nx2n+1(2n+1)!\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}

  3. cosx=1x22!+x44!x66!+=n=0(1)nx2n(2n)!\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}

  4. ln(1+x)=xx22+x33x44+=n=1(1)n1xnn(x<1)\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} \quad (|x| < 1)

11.4 Taylor Series for Two Variables

For f(x,y)f(x, y) about (a,b)(a, b):

f(x,y)=f(a,b)+fx(a,b)(xa)+fy(a,b)(yb)f(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b)
+12![fxx(a,b)(xa)2+2fxy(a,b)(xa)(yb)+fyy(a,b)(yb)2]++ \frac{1}{2!}[f_{xx}(a, b)(x-a)^2 + 2f_{xy}(a, b)(x-a)(y-b) + f_{yy}(a, b)(y-b)^2] + \cdots

12. Applications of Partial Derivatives

12.1 Tangent Plane to Surface

For surface z=f(x,y)z = f(x, y) at point (x0,y0,z0)(x_0, y_0, z_0):

Equation of tangent plane:

zz0=fx(x0,y0)(xx0)+fy(x0,y0)(yy0)z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)

Example: Find tangent plane to z=x2+y2z = x^2 + y^2 at (1,2,5)(1, 2, 5)

fx=2xfx(1,2)=2f_x = 2x \Rightarrow f_x(1, 2) = 2
fy=2yfy(1,2)=4f_y = 2y \Rightarrow f_y(1, 2) = 4

Tangent plane: z5=2(x1)+4(y2)z - 5 = 2(x - 1) + 4(y - 2)

z=2x2+4y8+5=2x+4y5z = 2x - 2 + 4y - 8 + 5 = 2x + 4y - 5

12.2 Linear Approximation

For f(x,y)f(x, y) near (a,b)(a, b):

L(x,y)=f(a,b)+fx(a,b)(xa)+fy(a,b)(yb)L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b)

Approximation: f(x,y)L(x,y)f(x, y) \approx L(x, y) for (x,y)(x, y) near (a,b)(a, b)

12.3 Optimization

To find local maxima/minima of f(x,y)f(x, y):

  1. Find critical points: Solve fx=0f_x = 0, fy=0f_y = 0

  2. Use Second Derivative Test: Let D=fxxfyy(fxy)2D = f_{xx}f_{yy} - (f_{xy})^2 at critical point (a,b)(a, b)

    • If D>0D > 0 and fxx>0f_{xx} > 0: Local minimum

    • If D>0D > 0 and fxx<0f_{xx} < 0: Local maximum

    • If D<0D < 0: Saddle point

    • If D=0D = 0: Test inconclusive

Example: Find and classify critical points of f(x,y)=x2+y22x4yf(x, y) = x^2 + y^2 - 2x - 4y

fx=2x2=0x=1f_x = 2x - 2 = 0 \Rightarrow x = 1
fy=2y4=0y=2f_y = 2y - 4 = 0 \Rightarrow y = 2

Critical point: (1,2)(1, 2)

Second derivatives: fxx=2f_{xx} = 2, fyy=2f_{yy} = 2, fxy=0f_{xy} = 0

D=(2)(2)(0)2=4>0D = (2)(2) - (0)^2 = 4 > 0

Since fxx=2>0f_{xx} = 2 > 0, (1,2)(1, 2) is local minimum.


13. Important Formulas Summary

13.1 Basic Differentiation Rules

  • Power rule: ddx(xn)=nxn1\frac{d}{dx}(x^n) = nx^{n-1}

  • Product rule: (fg)=fg+fg(fg)' = f'g + fg'

  • Quotient rule: (fg)=fgfgg2\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}

  • Chain rule: ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)

13.2 Common Derivatives

  • ddx(ex)=ex\frac{d}{dx}(e^x) = e^x

  • ddx(lnx)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}

  • ddx(sinx)=cosx\frac{d}{dx}(\sin x) = \cos x

  • ddx(cosx)=sinx\frac{d}{dx}(\cos x) = -\sin x

  • ddx(tan1x)=11+x2\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}

13.3 Partial Derivatives Notation

  • fx=fxf_x = \frac{\partial f}{\partial x}

  • fxy=2fyxf_{xy} = \frac{\partial^2 f}{\partial y \partial x}

  • Gradient: f=fx,fy\nabla f = \langle f_x, f_y \rangle

13.4 Chain Rule for Partials

zt=zxxt+zyyt\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}

14. Solved Examples

Example 1: Implicit Differentiation

Find dydx\frac{dy}{dx} for x2+xy+y2=3x^2 + xy + y^2 = 3

Solution:

Differentiate term by term:

2x+y+xdydx+2ydydx=02x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0
(x+2y)dydx=2xy(x + 2y)\frac{dy}{dx} = -2x - y
dydx=2xyx+2y\frac{dy}{dx} = \frac{-2x - y}{x + 2y}

Example 2: Logarithmic Differentiation

Find dydx\frac{dy}{dx} for y=(sinx)xy = (\sin x)^x

Solution:

Take ln\ln: lny=xln(sinx)\ln y = x \ln(\sin x)

Differentiate:

1ydydx=ln(sinx)+xcosxsinx\frac{1}{y}\frac{dy}{dx} = \ln(\sin x) + x \cdot \frac{\cos x}{\sin x}
dydx=y[ln(sinx)+xcotx]=(sinx)x[ln(sinx)+xcotx]\frac{dy}{dx} = y\left[\ln(\sin x) + x\cot x\right] = (\sin x)^x[\ln(\sin x) + x\cot x]

Example 3: Partial Derivatives

Find all second partials of f(x,y)=x3y+exyf(x, y) = x^3y + e^{xy}

Solution:

fx=3x2y+yexyf_x = 3x^2y + ye^{xy}
fy=x3+xexyf_y = x^3 + xe^{xy}
fxx=x(3x2y+yexy)=6xy+y2exyf_{xx} = \frac{\partial}{\partial x}(3x^2y + ye^{xy}) = 6xy + y^2e^{xy}
fyy=y(x3+xexy)=x2exyf_{yy} = \frac{\partial}{\partial y}(x^3 + xe^{xy}) = x^2e^{xy}
fxy=y(3x2y+yexy)=3x2+exy+xyexyf_{xy} = \frac{\partial}{\partial y}(3x^2y + ye^{xy}) = 3x^2 + e^{xy} + xye^{xy}
fyx=x(x3+xexy)=3x2+exy+xyexyf_{yx} = \frac{\partial}{\partial x}(x^3 + xe^{xy}) = 3x^2 + e^{xy} + xye^{xy}

Note fxy=fyxf_{xy} = f_{yx}


15. Common Mistakes and Exam Tips

15.1 Common Mistakes

  1. Chain rule errors: Forgetting to multiply by derivative of inner function

  2. Product/quotient rule: Misremembering formula signs

  3. Partial derivatives: Forgetting which variable is held constant

  4. Implicit differentiation: Forgetting to multiply by dydx\frac{dy}{dx} for yy terms

  5. Logarithmic differentiation: Forgetting final step of multiplying by yy

15.2 Problem-Solving Strategy

  1. Identify type: Which rule(s) apply? (product, quotient, chain, implicit, etc.)

  2. Proceed systematically: Write each step clearly

  3. Check work: Verify derivative makes sense

  4. Simplify: Final answer should be in simplest form

15.3 Quick Checks

  1. Derivative of constant = 0

  2. Chain rule: Always multiply by inside derivative

  3. Product rule: (fg)=fg+fg(fg)' = f'g + fg' (NOT fgf'g')

  4. Quotient rule: Denominator squared, minus sign in numerator

  5. Partial derivatives: Only differentiate with respect to one variable at a time

This comprehensive theory covers all aspects of ordinary and partial differentiation with detailed explanations and examples, providing complete preparation for the entrance examination.

Last updated