3.5 trig derivatives\n1. find the equation of the tangent line to (y = 5cos(x)) at (x=\frac{pi}{4}).

Explanation:

Step1: Find the derivative of the function

The derivative of $y = 5\cos(x)$ is $y'=- 5\sin(x)$ using the derivative formula for $\cos(x)$ which is $(\cos(x))'=-\sin(x)$ and the constant - multiple rule $(cf(x))' = cf'(x)$.

Step2: Evaluate the derivative at the given $x$ - value

Substitute $x = \frac{\pi}{4}$ into $y'$. So $y'\left(\frac{\pi}{4}\right)=-5\sin\left(\frac{\pi}{4}\right)=-5\times\frac{\sqrt{2}}{2}=-\frac{5\sqrt{2}}{2}$. This is the slope $m$ of the tangent line.

Step3: Find the $y$ - value of the function at the given $x$ - value

Substitute $x=\frac{\pi}{4}$ into $y = 5\cos(x)$. So $y\left(\frac{\pi}{4}\right)=5\cos\left(\frac{\pi}{4}\right)=5\times\frac{\sqrt{2}}{2}=\frac{5\sqrt{2}}{2}$.

Step4: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=\left(\frac{\pi}{4},\frac{5\sqrt{2}}{2}\right)$ and $m = -\frac{5\sqrt{2}}{2}$. [ \begin{align*} y-\frac{5\sqrt{2}}{2}&=-\frac{5\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)\ y-\frac{5\sqrt{2}}{2}&=-\frac{5\sqrt{2}}{2}x+\frac{5\sqrt{2}\pi}{8}\ y&=-\frac{5\sqrt{2}}{2}x+\frac{5\sqrt{2}\pi}{8}+\frac{5\sqrt{2}}{2} \end{align*} ]

Answer:

$y =-\frac{5\sqrt{2}}{2}x+\frac{5\sqrt{2}\pi}{8}+\frac{5\sqrt{2}}{2}$