07Gtrigonometricderivatives

toc = Derivatives of Trig Functions =

sin(x)
math . \qquad \text{For}\;f(x)=sin(x)\quad\;f'(x)=cos(x) \qquad. math

math . \qquad \text{For}\;f(x)=sin(kx),\quad\;f'(x)=kcos(kx) \qquad. math

cos(x)
math . \qquad \text{For}\;f(x)=cos(x),\quad\;f'(x)=-sin(x) \qquad. math

math . \qquad \text{For}\;f(x)=cos(kx),\quad\;f'(x)=-ksin(kx) \qquad. math

tan(x)
math . \qquad \text{For}\;f(x)=tan(x),\quad\;f'(x)=\dfrac{1}{cos^2(x)}=sec^2(x) \qquad. math

math . \qquad \text{For}\;f(x)=tan(kx),\quad\;f'(x)=\dfrac{k}{cos^2(kx)}=ksec^2(kx) \qquad. math

Note: math . \qquad \qquad \qquad \text{Multiply by } \; \dfrac{\pi}{180} \qquad. math
 * 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:


 * Example 1 **

Find the derivative of each of the following with respect to x. ... ... ... (a) y = sin (2x) ... ... ... (b) y = sin (xº) ... ... ... (c) y = sin 2 (2x) ... ... ... (d) y = cos (2x+1) 2 ... ... ... (e) y = tan (3x) ... ... ... (f) y = tan (3x 3 +1)

__Solution:__ (a) y = sin (2x)

math .\qquad\dfrac{dy}{dx}=2\cos(2x) \qquad. math

(b) y = sin (xº)

math . \qquad y = \sin(x^\circ) \qquad. \\ . \\ . \qquad y = \sin \Big( \dfrac{\pi x ^c}{180} \Big) \qquad. \\ . \\ . \qquad \dfrac {dy}{dx} = \dfrac {\pi}{180} \cos \Big( \dfrac{\pi x^c}{180} \Big) \qquad. math

(c) y = sin 2 (2x)

math .\quad\text{Let } u=\sin(2x) \text{ then } y=u^2 \qquad. \\ . \\ .\qquad\dfrac{du}{dx}=2\cos(2x) \text{ and } \dfrac{dy}{du}=2u \qquad. \\ . \\ .\qquad\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx} math

math .\qquad\dfrac{dy}{dx}=2u\times2\cos(2x) \qquad. \\ . \\ .\qquad\quad=4\sin(2x)\cos(2x) \qquad. \\ . \\ .\qquad\quad=2\sin(4x) \qquad \textit{using the double angle formula} \qquad. math

(d) y = cos (2x + 1) 2

math .\quad\text{Let } u=(2x+1)^2 \text{ then } y=\cos(u) \qquad. \\ . \\ .\qquad\dfrac{du}{dx}=2\times2\times(2x+1) \text{ and } \dfrac{dy}{du}=-\sin(u) \qquad. \\ . \\ .\qquad\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx} \qquad. math

math .\qquad\dfrac{dy}{dx}=-\sin(u)\times4(2x+1) \qquad. \\ . \\ .\qquad\quad=-4(2x+1)\sin(2x+1)^2 math

(e) y = tan (3x)

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

(f) y = tan (3x 3 + 1)

math .\quad\text{Let } u=3x^3+1 \text{ then } y=\tan(u) \qquad. \\ . \\ .\qquad\dfrac{du}{dx}=9x^2 \text{ and } \dfrac{dy}{du}=\dfrac{1}{\cos^2(u)} \qquad. \\ . \\ .\qquad\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx} math

math \\ .\qquad\dfrac{dy}{dx}=\dfrac{1}{\cos^2(u)}\times9x^2 \\ \\ .\qquad\quad =\dfrac{9x^2}{\cos^2(3x^3+1)} \\ \\ . \qquad \quad =9x^2\sec^2(3x^3+1) math

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

Note the answer for Example1 (d) and (e) 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 **
math \text{Find } \dfrac{dy}{dx} \; \text{ if }\;y= \tan \left( e^x \right) \qquad. math

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


 * __Solution:__**

math y= \tan \left( e^x \right) \qquad. math

math \\ .\quad \text{Let } u=e^x \; \text{ then } \; y= \tan(u) \qquad. \\ . \\ .\qquad\dfrac{du}{dx}=e^x \; \text{ and } \; \dfrac{dy}{du}=\dfrac{1}{ \cos^2(u)} \qquad. math

Use: math .\qquad\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx} \qquad. math

math \\ .\qquad\dfrac{dy}{dx}=\dfrac{1}{\cos^2(u)}\times e^x \\ \\ \\ .\qquad\quad=\dfrac{e^x}{\cos^2 \left( e^x \right) }=e^xsec^2 \left( e^x \right) math

Hence, at x = 0 math \\ . \quad \dfrac{dy}{dx} = \dfrac{e^0}{\cos^2 \left( e^0 \right) } \qquad. \\ . \\ . \qquad = \dfrac{1}{ \cos^2 (1) } \qquad. \\ . \\ . \qquad = \sec^2 (1) math

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