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Wize AP Calculus (BC) Textbook > Differentiation 2: Composite, Implicit, & Inverse Functions

Higher Order Derivatives

Derivatives of Inverse Trig FunctionsPrevious Section
Bonus: Logarithmic Differentiation (Important technique in calculus but not usually covered in AP Calculus AB or BC)Next Section
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Higher Order Derivatives

To get the second derivative, differentiate the first derivative. Then the third is the derivative of the second and so on.

The second derivative is denoted:

f′′(x)=y′′=d2fdx2=d2ydx2=d2dx2f(x)=ddx(dfdx)\boxed{\displaystyle f''(x) =y'' =\frac{d^2f}{dx^2} =\frac{d^2y}{dx^2} =\frac{d^2}{dx^2}f\left(x\right)=\frac{d}{dx}\left(\frac{df}{dx}\right)}

The nth derivative is denoted:

fn(x)=dnfdxn \boxed{\displaystyle f^n(x) =\frac{d^nf}{dx^n}\ }



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Example: Higher Order Derivatives

Find f′(π)+f′′(π)f'\left(\pi\right)+f''\left(\pi\right) if f(x)=x cos⁡x−sin⁡xf\left(x\right)=x\ \cos x-\sin x.

First Derivative
There's not much we can simplify with this expression, and since we have a product of functions, we need to use product rule.
f′(x)=(1)(cos⁡x)+(−sin⁡x)(x)−cos⁡xf'\left(x\right)=\left(1\right)\left(\cos x\right)+\left(-\sin x\right)\left(x\right)-\cos x
f′(x)=cos⁡x−x sin⁡x−cos⁡xf'\left(x\right)=\cos x-x\ \sin x-\cos x
f′(x)=−x sin⁡xf'\left(x\right)=-x\ \sin x

Second Derivative
Since we have a product of functions, we need the product rule again.
f′′(x)=(−1)(sin⁡x)+(cos⁡x)(−x)f''\left(x\right)=\left(-1\right)\left(\sin x\right)+\left(\cos x\right)\left(-x\right)
f′′(x)=−sin⁡x−xcos⁡xf''\left(x\right)=-\sin x-x\cos x

Substituting x=πx= \pi:
f′(π)+f′′(π)f'\left(\pi\right)+f''\left(\pi\right)
=−πsin⁡π+(−sin⁡π−πcos⁡π)=-\pi\sin\pi+\left(-\sin\pi-\pi\cos\pi\right)
=0+(0−π(−1))=0+\left(0-\pi\left(-1\right)\right)
=π=\pi
Find the third derivative of f(x)=x2+16xf(x) = x^2+16\sqrt{x}.
If f(x)=x10f\left(x\right)=x^{10} find f(9)(x)f^{\left(9\right)}\left(x\right).
(i.e. the 9th derivative)