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Derivatives of Trig Functions

We may often want to take derivatives involving the trigonometric functions. Rather than deriving them each time, it's much easier to simply remember the following derivatives.

Trig Derivatives

The derivatives of the six trigonometric functions are:

(sinx)=cosx          (cosx)=sinx(tanx)=sec2x      (secx)=secxtanx(cotx)=csc2x(cscx)=cscxcotx\begin{array}{c} \boxed{(\sin x)'=\cos x}\ \ \ \ \ \hspace{1 cm}\ \ \ \ \ \boxed{(\cos x)'=-\sin x } \\ \\ \boxed{(\tan x)'=\sec^{2} x} \ \ \ \ \ \ \qquad\boxed{(\sec x)'=\sec x\tan x} \\ \\ \boxed{ (\cot x)'=-\csc^{2}x}\qquad\boxed{(\csc x)'=-\csc x\cot x } \end{array}


Wize Tip
Notice how all the derivatives of "CO Functions" have a minus sign.

Wize Concept
The Product, Quotient, and Chain Rules still apply for Trig Derivatives!

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Example: Trig Derivatives


Find ddx(x2sinx)\displaystyle\frac{d}{dx}(x^2\sin{x}).

ddx(x2sinx)\displaystyle \frac{d}{dx}(x^2\sin{x})
=(ddxx2)×sinx+x2×(ddxsinx)\displaystyle =(\frac{d}{dx}x^2)\times\sin{x}+x^2\times(\frac{d}{dx}\sin{x})
=2xsinx+x2cosx\displaystyle =2x\sin{x}+x^2\cos{x}
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Example: Trig Derivatives

Find the derivative of f(x)=1+cosxxsinx\displaystyle f(x)=\frac{1+\cos x}{x-\sin x}


f(x)=y=1+cosxxsinx\displaystyle f(x)=y=\frac{1+\cos x}{x-\sin x}

Quotient rule:

y=sinx(xsinx)(1+cosx)(1cosx)(xsinx)2y'=\dfrac{-\sin x(x-\sin x)-(1+\cos x)(1-\cos x)}{(x-\sin x)^2}

=xsinx+sin2x1+cos2x(xsinx)2=\dfrac{-x\sin x+\sin ^2x-1+\cos ^2x}{(x-\sin x)^2}

=xsinx(xsinx)2=-\dfrac{x\sin x}{(x-\sin x)^2}
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Example: Trig Derivatives

Find the derivative of the following function

y=sinxy=\sin\sqrt{x}

The outside function is f(x)=sin(x)f(x)=\sin (x) and the inside function is g(x)=x=x12g(x)=\sqrt{x}=x^{-\frac{1}{2}}.

The chain rule tells us that the derivative is the derivative of the outside function (keeping the inside the same), times the derivative of the inside function:
y=f(g(x))×g(x)y'=f'\left(g\left(x\right)\right)\times g'\left(x\right)
y=cos(x)×12x12\displaystyle y'=\cos\left(\sqrt{x}\right)\times\frac{1}{2}x^{-\frac{1}{2}}

Simplifying:
y=cos(x)2x\displaystyle y'=\frac{\cos\left(\sqrt{x}\right)}{2\sqrt{x}}
Find the derivative of

f(θ)=(sinθcosθ)secθ\displaystyle f(\theta)=(\sin\theta-\cos\theta)\sec\theta



Find y(23)y^{(23)} , the 23rd derivatives, given y=sinx\displaystyle y=\sin x.

Extra Practice