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Derivative of Trig Functions

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Derivative of Trigonometric Functions
The derivative of the function f(x)=sin⁡xf\left(x\right)=\sin xf(x)=sinx is f′(x)=cos⁡xf'\left(x\right)=\cos xf′(x)=cosx
(i.e. ddx(sin⁡x)=cos⁡x\frac{d}{dx}\left(\sin x\right)=\cos xdxd​(sinx)=cosx)

The derivative of the function f(x)=cos⁡xf\left(x\right)=\cos xf(x)=cosx is f′(x)=−sin⁡xf'\left(x\right)=-\sin xf′(x)=−sinx
(i.e. ddx(cos⁡x)=−sin⁡x\frac{d}{dx}\left(\cos x\right)=-\sin xdxd​(cosx)=−sinx)

The derivative of the function f(x)=tan⁡xf\left(x\right)=\tan xf(x)=tanx is f′(x)=sec⁡2xf'\left(x\right)=\sec^2xf′(x)=sec2x
(i.e. ddx(tan⁡x)=sec⁡2x\frac{d}{dx}\left(\tan x\right)=\sec^2xdxd​(tanx)=sec2x)

Wize Tip
The product, quotient, and chain rules all still apply with the trigonometric functions.

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Example
Find the derivatives of the following functions.
a) y=sin⁡x−xcos⁡xy=\sin x-x\cos xy=sinx−xcosx

Using product rule on the second term:
y′=(sin⁡x)′−[(x)′(cos⁡x)+(x)(cos⁡x)′]y'=\left(\sin x\right)'-\left[\left(x\right)'\left(\cos x\right)+\left(x\right)\left(\cos x\right)'\right]y′=(sinx)′−[(x)′(cosx)+(x)(cosx)′]
y′=cos⁡x−[(1(cos⁡x)+(x)(−sin⁡x))]y'=\cos x-\left[\left(1\left(\cos x\right)+\left(x\right)\left(-\sin x\right)\right)\right]y′=cosx−[(1(cosx)+(x)(−sinx))]
y′=cos⁡x−cos⁡x−xsin⁡xy'=\cos x-\cos x-x\sin xy′=cosx−cosx−xsinx
y′=−xsin⁡xy'=-x\sin xy′=−xsinx

b) y=sin⁡xcos⁡xy=\sin x\cos xy=sinxcosx

Using product rule:
y′=(sin⁡x)′(cos⁡x)+(sin⁡x)(cos⁡x)′y'=\left(\sin x\right)'\left(\cos x\right)+\left(\sin x\right)\left(\cos x\right)'y′=(sinx)′(cosx)+(sinx)(cosx)′
y′=(cos⁡x)(cos⁡x)+(sin⁡x)(−sin⁡x)y'=\left(\cos x\right)\left(\cos x\right)+\left(\sin x\right)\left(-\sin x\right)y′=(cosx)(cosx)+(sinx)(−sinx)
y′=cos⁡2x−sin⁡2xy'=\cos^2x-\sin^2xy′=cos2x−sin2x

c) y=cos⁡(x3)y=\cos\left(x^3\right)y=cos(x3)

Using chain rule:
y′=(cos⁡...)′×(x3)′y'=\left(\cos...\right)'\times\left(x^3\right)'y′=(cos...)′×(x3)′
y′=−sin⁡(x3)×3x2y'=-\sin\left(x^3\right)\times3x^2y′=−sin(x3)×3x2
y′=−3x2sin⁡(x3)y'=-3x^2\sin\left(x^3\right)y′=−3x2sin(x3)

d) y=cos⁡3xy=\cos^3xy=cos3x (i.e. y=(cos⁡x)3y=\left(\cos x\right)^3y=(cosx)3)

Using chain rule:
y′=[(...)3]′×(cos⁡x)′y'=\left[\left(...\right)^3\right]'\times\left(\cos x\right)'y′=[(...)3]′×(cosx)′
y′=3(cos⁡x)2×(−sin⁡x)y'=3\left(\cos x\right)^2\times\left(-\sin x\right)y′=3(cosx)2×(−sinx)
y′=−3sin⁡xcos⁡2xy'=-3\sin x\cos^2xy′=−3sinxcos2x

e) y=tan⁡(sin⁡x)y=\tan\left(\sin x\right)y=tan(sinx)

Using chain rule:
y′=[tan⁡(...)]′×(sin⁡x)′y'=\left[\tan\left(...\right)\right]'\times\left(\sin x\right)'y′=[tan(...)]′×(sinx)′
y′=sec⁡2(sin⁡x)×cos⁡xy'=\sec^2\left(\sin x\right)\times\cos xy′=sec2(sinx)×cosx

Practice: Derivative of Trigonometric Functions
Find the derivative of the following functions:
a) f(x)=x2sin⁡xf\left(x\right)=x^2\sin xf(x)=x2sinx

b) f(x)=sin⁡(x3)f\left(x\right)=\sin\left(x^3\right)f(x)=sin(x3)

c) f(x)=sin⁡3xf\left(x\right)=\sin^3xf(x)=sin3x

d) f(x)=etan⁡xf\left(x\right)=e^{\sqrt{\tan x}}f(x)=etanx​

a) Enter the derivative

f'(x)=

b) Enter the derivative

f'(x)=

c) Enter the derivative

f'(x)=

d) Enter the derivative

f'(x)=

I don't know

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Example: Equation of Tangent Line
Find the equation of the tangent line to the curve y=2xcos⁡xy=\frac{2^x}{\cos x}y=cosx2x​ at the point (0,1)\left(0,1\right)(0,1).

1. We have a point on the tangent -- (0,1)\left(0,1\right)(0,1)

2. The slope of the tangent line at that point is the derivative of the curve at that point:
y′=(derivative of 1st)(2nd)−(derivative of 2nd)(1st)(2nd)2y'=
\frac{\color{blue}(\text{derivative of 1st})
\color{orange}(\text{2nd})
\color{black}-
\color{red}(\text{derivative of 2nd})
\color{green}(\text{1st})}
{\color{purple}(\text{2nd})^2}y′=(2nd)2(derivative of 1st)(2nd)−(derivative of 2nd)(1st)​
y′=(2x⋅ln⁡2)(cos⁡x)+(−sin⁡x)(2x)cos⁡2xy'=
\frac{\color{blue}\left(2^x\cdot\ln2\right)\color{orange}\left(\cos x\right)
\color{black}+
\color{red}\left(-\sin x\right)
\color{green}\left(2^x\right)}
{\color{purple}\cos^2x}y′=cos2x(2x⋅ln2)(cosx)+(−sinx)(2x)​
Now we sub in the value x=0:
y′(0)=(20⋅ln⁡2)(cos⁡0)+(−sin⁡0)(20)cos⁡20y'(0)=\frac{
\color{blue}{\left(2^0\cdot\ln2\right)}
\color{orange}{\left(\cos 0\right)}
\color{black}{+}
\color{red}{\left(-\sin 0\right)}
\color{green}{\left(2^0\right)}}
{\color{purple}{\cos^20}}y′(0)=cos20(20⋅ln2)(cos0)+(−sin0)(20)​
y′(0)=ln⁡2−01y'(0)=\frac{\ln2-0}{1}y′(0)=1ln2−0​
y′(0)=ln⁡2y'(0)=\ln2y′(0)=ln2
So, the slope of the tangent line at that point is ln⁡2\ln2ln2

3. Substitute all this into the line equation:
y−y1=m(x−x1)y-y_1=m\left(x-x_1\right)y−y1​=m(x−x1​)
y−1=ln⁡2(x−0)y-1=\ln2\left(x-0\right)y−1=ln2(x−0)
y−1=(ln⁡2)(x)y-1=\left(\ln2\right)\left(x\right)y−1=(ln2)(x)

If this was a multiple choice question, we would check the answer options at this point, we might have to rearrange to get the equation of the tangent line y=(ln⁡2)x+1y=\left(\ln2\right)x+1y=(ln2)x+1

Practice: Equation of Tangent Line
Find the equation of the tangent line to the graph y=sin⁡x+3x2cos⁡xy=\sin x+3x^2\cos xy=sinx+3x2cosx at the point where x=π2x=\frac{\pi}{2}x=2π​.

Enter the equation of the tangent line in the form

y=mx+b

I don't know

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Example: Higher Order Derivative
Find f′(π)+f′′(π)f'\left(\pi\right)+f''\left(\pi\right)f′(π)+f′′(π) if f(x)=x cos⁡x−sin⁡xf\left(x\right)=x\ \cos x-\sin xf(x)=x cosx−sinx.

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 xf′(x)=(1)(cosx)+(−sinx)(x)−cosx
f′(x)=cos⁡x−x sin⁡x−cos⁡xf'\left(x\right)=\cos x-x\ \sin x-\cos xf′(x)=cosx−x sinx−cosx
f′(x)=−x sin⁡xf'\left(x\right)=-x\ \sin xf′(x)=−x sinx

Second Derivative
Since we have a produce of funcions, 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)=(−1)(sinx)+(cosx)(−x)
f′′(x)=−sin⁡x−xcos⁡xf''\left(x\right)=-\sin x-x\cos xf′′(x)=−sinx−xcosx

Substituting the value for x:
f′(π)+f′′(π)f'\left(\pi\right)+f''\left(\pi\right)f′(π)+f′′(π)
=−πsin⁡π+(−sin⁡π−πcos⁡π)=-\pi\sin\pi+\left(-\sin\pi-\pi\cos\pi\right)=−πsinπ+(−sinπ−πcosπ)
=0+(0−π(−1))=0+\left(0-\pi\left(-1\right)\right)=0+(0−π(−1))
=π=\pi=π

Practice: Higher Order Derivative
If f(x)=sin⁡(2x)f\left(x\right)=\sin\left(2x\right)f(x)=sin(2x), find f(21)(π2)f^{\left(21\right)}\left(\frac{\pi}{2}\right)f(21)(2π​). 
(i.e. find the 21st derivative at the point π2\frac{\pi}{2}2π​)

Practice: Derivative of Trig Functions
A differential equation is an equation that has some derivatives in it (i.e. it may contain y, y′, y′′, etc.y,\ y',\ y'',\ etc.y, y′, y′′, etc.)

If y=sin⁡kx+cos⁡kxy=\sin kx+\cos kxy=sinkx+coskx, determine the value(s) of the constant kkk such that y′′+y′+(k+k2)y=0y''+y'+\left(k+k^2\right)y=0y′′+y′+(k+k2)y=0 is true for all values of xxx.

k=0k=0k=0 only

k=0, π2k=0,\ \frac{\pi}{2}k=0, 2π​ only

k=0. −π2, π2k=0.\ -\frac{\pi}{2},\ \frac{\pi}{2}k=0. −2π​, 2π​ only

k=1k=1k=1 only

k=±1k=\pm1k=±1 only

I don't know