homework5: problem 36 (1 point) find an equation of the tangent line to the curve y = sin(7x)+cos(8x) at the…

Explanation:

Step1: Find the derivative of y

Using the chain - rule, if $y = \sin(7x)+\cos(8x)$, then $y^\prime=\frac{d}{dx}(\sin(7x))+\frac{d}{dx}(\cos(8x))$. The derivative of $\sin(u)$ with respect to $x$ is $\cos(u)\cdot u^\prime$ and the derivative of $\cos(u)$ with respect to $x$ is $-\sin(u)\cdot u^\prime$. Here $u = 7x$ for the first term and $u = 8x$ for the second term. So $y^\prime=7\cos(7x)-8\sin(8x)$.

Step2: Evaluate the derivative at $x = \frac{\pi}{6}$

$y^\prime\left(\frac{\pi}{6}\right)=7\cos\left(7\times\frac{\pi}{6}\right)-8\sin\left(8\times\frac{\pi}{6}\right)$. $\cos\left(\frac{7\pi}{6}\right)=-\frac{\sqrt{3}}{2}$ and $\sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}$. $y^\prime\left(\frac{\pi}{6}\right)=7\times\left(-\frac{\sqrt{3}}{2}\right)-8\times\left(-\frac{\sqrt{3}}{2}\right)=\frac{-7\sqrt{3}+8\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$.

Step3: Find the value of y at $x=\frac{\pi}{6}$

$y\left(\frac{\pi}{6}\right)=\sin\left(7\times\frac{\pi}{6}\right)+\cos\left(8\times\frac{\pi}{6}\right)$. $\sin\left(\frac{7\pi}{6}\right)=-\frac{1}{2}$ and $\cos\left(\frac{4\pi}{3}\right)=-\frac{1}{2}$. So $y\left(\frac{\pi}{6}\right)=-\frac{1}{2}-\frac{1}{2}=-1$.

Step4: Use the point - slope form $y - y_1=m(x - x_1)$

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=\left(\frac{\pi}{6},-1\right)$ and $m = \frac{\sqrt{3}}{2}$. $y+1=\frac{\sqrt{3}}{2}\left(x-\frac{\pi}{6}\right)$. $y=\frac{\sqrt{3}}{2}x-\frac{\sqrt{3}\pi}{12}-1$.

Answer:

$y=\frac{\sqrt{3}}{2}x-\frac{\sqrt{3}\pi}{12}-1$