Advanced Techniques for Computing Derivatives

Advanced Techniques for Computing Derivatives

Once the basic differentiation rules and the basic computation methods are mastered, we need to learn some advanced techniques to handle more complicated differentiation problems.

Higher-Order Derivatives

Second Derivative

Second Derivative

f′′(x)=(f′(x))′f''(x) = (f'(x))'

Physical meaning: acceleration is the derivative of velocity, that is, the second derivative of displacement.

Examples:

  • f(x)=x2f(x) = x^2, f′(x)=2xf'(x) = 2x, f′′(x)=2f''(x) = 2
  • f(x)=sin⁡xf(x) = \sin x, f′(x)=cos⁡xf'(x) = \cos x, f′′(x)=−sin⁡xf''(x) = -\sin x

n-th Derivative

n-th Derivative

f(n)(x)=(f(n−1)(x))′f^{(n)}(x) = (f^{(n-1)}(x))'

Higher-order derivatives of common functions:

  • (xn)(n)=n!(x^n)^{(n)} = n!
  • (ex)(n)=ex(e^x)^{(n)} = e^x
  • (sin⁡x)(n)=sin⁡(x+nπ2)(\sin x)^{(n)} = \sin(x + \frac{n\pi}{2})
  • (cos⁡x)(n)=cos⁡(x+nπ2)(\cos x)^{(n)} = \cos(x + \frac{n\pi}{2})

Implicit Differentiation

The Basic Method

Steps:

  1. Differentiate both sides of the equation simultaneously
  2. Solve for dydx\frac{dy}{dx}, treating it as an unknown
  3. Rearrange to get an expression for dydx\frac{dy}{dx}

Example: Find the derivative of x2+y2=1x^2 + y^2 = 1.

Solution:

  1. Differentiate both sides: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0
  2. Rearrange: 2ydydx=−2x2y\frac{dy}{dx} = -2x
  3. Obtain: dydx=−xy\frac{dy}{dx} = -\frac{x}{y}

Complicated Implicit Functions

Example: Find the derivative of x3+y3=3xyx^3 + y^3 = 3xy.

Solution:

  1. Differentiate both sides: 3x2+3y2dydx=3y+3xdydx3x^2 + 3y^2\frac{dy}{dx} = 3y + 3x\frac{dy}{dx}
  2. Rearrange: 3y2dydx−3xdydx=3y−3x23y^2\frac{dy}{dx} - 3x\frac{dy}{dx} = 3y - 3x^2
  3. Factor out the common term: (3y2−3x)dydx=3y−3x2(3y^2 - 3x)\frac{dy}{dx} = 3y - 3x^2
  4. Obtain: dydx=y−x2y2−x\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}

Differentiation of Parametric Equations

The Basic Formula

Parametric Differentiation Formula

dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

Explanation: the derivative of a parametric curve equals the derivative of yy with respect to the parameter divided by the derivative of xx with respect to the parameter.

Example: Find the derivative of the parametric equations x=t2x = t^2, y=t3y = t^3.

Solution:

  • dxdt=2t\frac{dx}{dt} = 2t
  • dydt=3t2\frac{dy}{dt} = 3t^2
  • dydx=3t22t=3t2\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}

Higher-Order Derivatives

Second derivative: d2ydx2=ddx(dydx)=ddt(dydx)⋅dtdx\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx}

Example: Find the second derivative of the parametric equations x=cos⁡tx = \cos t, y=sin⁡ty = \sin t.

Solution:

  1. First derivative: dydx=cos⁡t−sin⁡t=−cot⁡t\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t
  2. Second derivative: d2ydx2=ddt(−cot⁡t)⋅1−sin⁡t=csc⁡2t⋅1−sin⁡t=−csc⁡3t\frac{d^2y}{dx^2} = \frac{d}{dt}(-\cot t) \cdot \frac{1}{-\sin t} = \csc^2 t \cdot \frac{1}{-\sin t} = -\csc^3 t

Logarithmic Differentiation

When to Use It

Logarithmic differentiation is useful in the following situations:

  • complicated power-exponential functions
  • products or quotients of several functions
  • expressions that contain both exponentials and logarithms

The Basic Steps

  1. Take the logarithm of the function
  2. Differentiate both sides of the equation
  3. Solve for dydx\frac{dy}{dx}

Examples

Example 1: Find the derivative of y=xxy = x^x.

Solution:

  1. Take the logarithm: ln⁡y=xln⁡x\ln y = x \ln x
  2. Differentiate: 1ydydx=ln⁡x+1\frac{1}{y}\frac{dy}{dx} = \ln x + 1
  3. Solve: dydx=y(ln⁡x+1)=xx(ln⁡x+1)\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)

Example 2: Find the derivative of y=(x+1)2(x−2)3(x+3)4y = \frac{(x+1)^2(x-2)^3}{(x+3)^4}.

Solution:

  1. Take the logarithm: ln⁡y=2ln⁡(x+1)+3ln⁡(x−2)−4ln⁡(x+3)\ln y = 2\ln(x+1) + 3\ln(x-2) - 4\ln(x+3)
  2. Differentiate: 1ydydx=2x+1+3x−2−4x+3\frac{1}{y}\frac{dy}{dx} = \frac{2}{x+1} + \frac{3}{x-2} - \frac{4}{x+3}
  3. Solve: dydx=y(2x+1+3x−2−4x+3)\frac{dy}{dx} = y\left(\frac{2}{x+1} + \frac{3}{x-2} - \frac{4}{x+3}\right)

Differentiation of Piecewise Functions

The Basic Method

Method:

  1. Differentiate each piece separately
  2. Check differentiability at the breakpoints

Example

Example: f(x)={x2,x≥0−x2,x<0f(x) = \begin{cases} x^2, & x \geq 0 \\ -x^2, & x < 0 \end{cases}

Solution:

  • When x>0x > 0: f′(x)=2xf'(x) = 2x
  • When x<0x < 0: f′(x)=−2xf'(x) = -2x
  • At x=0x = 0: f′(0)=0f'(0) = 0

Advanced Computation Techniques

1. Simplification

Technique: simplify the function before differentiating it

Example:

  • Original function: f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}
  • Simplified: f(x)=x+1f(x) = x + 1 (when x≠1x \neq 1)
  • Derivative: f′(x)=1f'(x) = 1

2. Exploiting Symmetry

Technique: use the symmetry of the function to simplify the computation

Example: For an even function f(x)=f(−x)f(x) = f(-x):

  • f′(x)=−f′(−x)f'(x) = -f'(-x)
  • At x=0x = 0, f′(0)=0f'(0) = 0

For an odd function f(x)=−f(−x)f(x) = -f(-x):

  • f′(x)=f′(−x)f'(x) = f'(-x)
  • the derivative is an even function

3. Decomposing Composite Functions

Technique: decompose a complicated function into a composition of simple ones

Example: f(x)=sin⁡(ex2)f(x) = \sin(e^{x^2})

Decomposition:

  • u=x2u = x^2
  • v=euv = e^u
  • w=sin⁡vw = \sin v
  • f(x)=wf(x) = w

Differentiation:

  • dwdv=cos⁡v=cos⁡(ex2)\frac{dw}{dv} = \cos v = \cos(e^{x^2})
  • dvdu=eu=ex2\frac{dv}{du} = e^u = e^{x^2}
  • dudx=2x\frac{du}{dx} = 2x
  • f′(x)=cos⁡(ex2)⋅ex2⋅2x=2xex2cos⁡(ex2)f'(x) = \cos(e^{x^2}) \cdot e^{x^2} \cdot 2x = 2xe^{x^2}\cos(e^{x^2})

Common Errors and Precautions

1. Mistakes in Implicit Differentiation

Mistake: forgetting to apply the chain rule when differentiating yy Correct: when differentiating a term that contains yy, multiply by dydx\frac{dy}{dx}

Example:

  • Mistake: (y2)′=2y(y^2)' = 2y
  • Correct: (y2)′=2ydydx(y^2)' = 2y\frac{dy}{dx}

2. Mistakes in Differentiation of Parametric Equations

Mistake: differentiating xx and yy directly Correct: differentiate with respect to the parameter first, then use the formula

3. Mistakes in Logarithmic Differentiation

Mistake: forgetting to multiply by the original function at the end Correct: dydx=y⋅ddx(ln⁡y)\frac{dy}{dx} = y \cdot \frac{d}{dx}(\ln y)


Exercises

Exercise 1

Find the derivative of the implicit function x2+xy+y2=1x^2 + xy + y^2 = 1.

derivative calculationdifferentiation rules

Approach: Use implicit differentiation.

Steps:

  1. Differentiate both sides: 2x+y+xdydx+2ydydx=02x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0

  2. Rearrange: (x+2y)dydx=−(2x+y)(x + 2y)\frac{dy}{dx} = -(2x + y)

  3. Solve: dydx=−2x+yx+2y\frac{dy}{dx} = -\frac{2x + y}{x + 2y}

Answer: dydx=−2x+yx+2y\frac{dy}{dx} = -\frac{2x + y}{x + 2y}

Exercise 2

Find the derivative of the parametric equations x=t3x = t^3, y=t2y = t^2.

derivative calculationdifferentiation rules

Approach: Use the parametric differentiation formula.

Steps:

  1. Compute dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}: dxdt=3t2\frac{dx}{dt} = 3t^2 dydt=2t\frac{dy}{dt} = 2t

  2. Apply the formula: dydx=dydtdxdt=2t3t2=23t\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t}{3t^2} = \frac{2}{3t}

Answer: dydx=23t\frac{dy}{dx} = \frac{2}{3t}

Exercise 3

Find the derivative of y=xsin⁡xy = x^{\sin x} using logarithmic differentiation.

derivative calculationdifferentiation rules

Approach: Use logarithmic differentiation.

Steps:

  1. Take the logarithm: ln⁡y=sin⁡x⋅ln⁡x\ln y = \sin x \cdot \ln x

  2. Differentiate both sides: 1ydydx=cos⁡x⋅ln⁡x+sin⁡x⋅1x\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}

  3. Solve: dydx=y(cos⁡x⋅ln⁡x+sin⁡xx)=xsin⁡x(cos⁡x⋅ln⁡x+sin⁡xx)\frac{dy}{dx} = y(\cos x \cdot \ln x + \frac{\sin x}{x}) = x^{\sin x}(\cos x \cdot \ln x + \frac{\sin x}{x})

Answer: dydx=xsin⁡x(cos⁡x⋅ln⁡x+sin⁡xx)\frac{dy}{dx} = x^{\sin x}(\cos x \cdot \ln x + \frac{\sin x}{x})

Exercise 4

Find the derivative of f(x)={x2,x≥0−x2,x<0f(x) = \begin{cases} x^2, & x \geq 0 \\ -x^2, & x < 0 \end{cases}.

derivative calculationdifferentiation rules

Approach: Differentiate a piecewise function.

Steps:

  1. When x>0x > 0: f′(x)=2xf'(x) = 2x

  2. When x<0x < 0: f′(x)=−2xf'(x) = -2x

  3. Check differentiability at x=0x = 0:

    • Left derivative: lim⁡h→0−f(h)−f(0)h=lim⁡h→0−−h2h=0\lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^2}{h} = 0
    • Right derivative: lim⁡h→0+f(h)−f(0)h=lim⁡h→0+h2h=0\lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2}{h} = 0
    • Since the left derivative equals the right derivative, f′(0)=0f'(0) = 0

Answer: f′(x)={2x,x>0−2x,x<00,x=0f'(x) = \begin{cases} 2x, & x > 0 \\ -2x, & x < 0 \\ 0, & x = 0 \end{cases}


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
f(n)(x)f^{(n)}(x)Mathematical symbolf n-th derivative of xThe n-th derivative, defined by f(n)(x)=(f(n−1)(x))′f^{(n)}(x) = (f^{(n-1)}(x))'
n!n!Mathematical symboln factorialThe constant value of the n-th derivative of the power function xnx^n
nπ2\frac{n\pi}{2}Mathematical symboln pi over twoThe phase shift in the n-th derivatives of sine and cosine
f′′(x)f''(x)Mathematical symbolf double prime of xThe second derivative, read physically as the acceleration of a moving point
dydx\frac{dy}{dx}Mathematical symbold y over d xThe derivative sought by the implicit, parametric and logarithmic techniques
dxdt\frac{dx}{dt}Mathematical symbold x over d tThe derivative of the parametric x-component with respect to the parameter
dtdx\frac{dt}{dx}Mathematical symbold t over d xThe factor that turns a derivative with respect to tt into one with respect to xx
ln⁡y\ln yMathematical symbolnatural log of yThe logarithm taken in logarithmic differentiation, which leaves 1ydydx\frac{1}{y}\frac{dy}{dx} on the left
u,v,wu, v, wMathematical symbolu, v, wThe three intermediate variables of the decomposition f(x)=sin⁡(ex2)f(x) = \sin(e^{x^2})
dwdv\frac{dw}{dv}Mathematical symbold w over d vThe outermost layer factor in the decomposed derivative of sin⁡(ex2)\sin(e^{x^2})

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