05 8trigfuns

# Derivatives of Trig Functions

### sin(x)

… … For $f(x)=\sin(x)\quad\;f'(x)=\cos(x)$

… … For $f(x)=\sin(kx),\quad\;f'(x)=k\cos(kx)$

### cos(x)

… … For $f(x)=\cos(x),\quad\;f'(x)=-\sin(x)$

… … For $f(x)=\cos(kx),\quad\;f'(x)=-k\sin(kx)$

### tan(x)

… … For $f(x)=\tan(x),\quad\;f'(x)=\dfrac{1}{\cos^2(x)}=\sec^2(x)$

… … For $f(x)=\tan(kx),\quad\;f'(x)=\dfrac{k}{\cos^2(kx)}=k\sec^2(kx)$

.

Note:

• Calculus with trigonometry is ALWAYS done using radians.
• If you are given a value that is in degrees, you will need to convert to radians before performing calculus.
• Remember that the rule for converting from degrees to radians is multiply by $\dfrac{\pi}{180}$

.

#### Example 1

Find the derivative of each of the following with respect to x.

a)$y = \sin (2x)$

b)$y = \sin (x^\circ)$

c)$y = \sin^2 (2x)$

d)$y = \cos \big( (2x+1)^2 \big)$

e)$y = \tan (3x)$

f)$y = \tan (3x^3+1)$

Solution:

a)$y = \sin (2x)$

… … … $\dfrac{dy}{dx}=2\cos(2x)$
.

b)$y = \sin (x^\circ)$

… … … $y = \sin \Big( \dfrac{\pi x ^c}{180} \Big)$

… … … $\dfrac{dy}{dx} = \dfrac {\pi}{180} \cos \Big( \dfrac{\pi x^c}{180} \Big)$
.

c)$y = \sin^2 (2x)$

… … … Let $u=\sin(2x) \text{ then } y=u^2$

… … … $\dfrac{du}{dx}=2\cos(2x) \text{ and } \dfrac{dy}{du}=2u$

… … … $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$

… … … $\dfrac{dy}{dx}=2u\times2\cos(2x)$

… … … … $=4\sin(2x)\cos(2x)$

… … … … $=2\sin(4x)$ … … using the double angle formula
.

d)$y = \cos \big( (2x+1)^2 \big)$

… … … Let $u=(2x+1)^2 \text{ then } y=\cos(u)$

… … … $\dfrac{du}{dx}=2\times2\times(2x+1) \text{ and } \dfrac{dy}{du}=-\sin(u)$

… … … $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$

… … … $\dfrac{dy}{dx}=-\sin(u)\times4(2x+1)$

… … … … $=-4(2x+1)\sin\big((2x+1)^2\big)$
.

e)$y = \tan (3x)$

… … … $\dfrac{dy}{dx}=\dfrac{3}{ \cos^2(3x)}=3\sec^2(3x)$
.

f)$y = \tan (3x^3+1)$

… … … Let $u=3x^3+1 \text{ then } y=\tan(u)$

… … … $\dfrac{du}{dx}=9x^2 \text{ and } \dfrac{dy}{du}=\dfrac{1}{\cos^2(u)}$

… … … $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$

… … … $\dfrac{dy}{dx}=\dfrac{1}{\cos^2(u)}\times9x^2$

… … … … $=\dfrac{9x^2}{\cos^2(3x^3+1)}$

… … … … $=9x^2\sec^2(3x^3+1)$

.

When using the CAS calculator to differentiate trigonometric functions, ensure the $RADIAN$ mode is on.

.

Note the answer for (1d) and (1e) when usng the CAS calculator
When finding the derivative of the tangent function the CAS calculator's answer has been re-written using the identity: $\sec^2(x) = \tan^2 (x) + 1$.

.

#### Example 2

Find $\dfrac{dy}{dx} \; \text{ if }\;y= \tan \left( e^x \right)$

Hence find the gradient of a tangent at the point $x = 0$

Solution:

… … $y= \tan \left( e^x \right)$

… … Let $u=e^x \; \text{ then } \; y= \tan(u)$

… … $\dfrac{du}{dx}=e^x \; \text{ and } \; \dfrac{dy}{du}=\dfrac{1}{ \cos^2(u)}$

… … Use:

… … $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$

… … $\dfrac{dy}{dx}=\dfrac{1}{\cos^2(u)}\times e^x$

… … … $=\dfrac{e^x}{\cos^2 \left( e^x \right) }$

… … … $=e^x\sec^2 \left( e^x \right)$

… … Hence, at $x = 0$

… … $\dfrac{dy}{dx} = \dfrac{e^0}{\cos^2 \left( e^0 \right) }$

… … … $= \dfrac{1}{ \cos^2 (1) }$

… … … $= \sec^2 (1)$

.