3.5 – Trigonometric Derivatives

The Sine Function

$$\frac{d}{dx}\sin x=\cos x$$

The Cosine Function

$$\frac{d}{dx}\cos x=-\sin x$$

The cyclic nature of the sine function means that we eventually return to $\sin x$ with enough derivations.

$$y=\sin x$$ $$y^{\prime }=\cos x$$ $$y^{\prime \prime }=-\sin x$$ $$y^{\prime \prime \prime }=-\cos x$$ $$y^{\prime \prime \prime \prime }=\sin x$$

Tangent and Cotangent

$$\frac{d}{dx}\tan x={\sec }^{2}x$$ $$\frac{d}{dx}\cot x=-{\csc }^{2}x$$

Secant and Cosecant

$$\frac{d}{dx}\sec x=\sec x\tan x$$ $$\frac{d}{dx}\csc x=-\csc x\cot x$$

Examples

Find the equation of the normal line to $f(x)=\frac{\tan x}{x}$ at $x=3$.

Solution

The slope of the normal line is equal to the negative reciprocal of the tangent slope, which is the first derivative.

$$m_{\text{tan}}=f^{\prime }(x)=\frac{x{\sec }^{2}x-\tan x}{x^2}$$ $$m_{\text{tan}}=\frac{3{\sec }^{2}3-\tan 3}{3^2}=0.356$$ $$m_{\text{norm}}=-\frac{1}{m_{\text{tan}}}=-\frac{1}{0.356}$$ $$m_{\text{norm}}=-2.809$$

Using the point-slope form at $(x,f(x))$:

$$x=3,f(x)=-0.048$$ $$y=-2.809(x-3)+0.048$$

Find the second derivative of $f(x)={\sin }^{2}x$.

Solution (without the chain rule)

Using the product rule to find the first derivative:

$$f^{\prime }(x)=\sin x\cos x+\sin x\cos x=2\sin x\cos x$$

Taking out the constant multiple and applying the product rule again:

$$f^{\prime \prime }(x)=2[\frac{d}{dx}\sin x\cos x]$$ $$=2[{\cos }^{2}x-{\sin }^{2}x]$$ $$=2\cos (2x)$$

Solution (chain rule)

Since ${\sin }^{2}x$ takes the form of $f(g(x))$, we can use the chain rule.

The outside function $f(x)$ is $x^2$. The inside function $g(x)$ is $\sin x$. Therefore, $f^{\prime }(x)=2x$ and $g^{\prime }(x)=\cos x$.

Substituting our functions into the chain rule $f^{\prime }(g(x))\cdot g^{\prime }(x)$:

$$\frac{d}{dx}{\sin }^{2}x=2\sin x\cos x=\sin (2x)$$

The chain rule can be applied again with the $\sin x$ as the outside function and $2x$ as the inside function. Therefore, $f^{\prime }(x)$ is $\cos x$ and $g^{\prime }(x)$ is $2$.

$$\frac{d^2}{d{x}^2}{\sin }^{2}x=\cos (2x)\cdot 2$$ $$=2\cos (2x)$$