I came across this really confusing tangent line question.
Given the following prompt: "Let $f$ be an increasing function with $f(0)=3$. The derivative of $f$ is given by $f'(x)=\cos(\pi x)+x^4+6$. Write an equation for the line tangent to the graph of $y=(f(x))^2$ at $x=0$."
I've been trying to work through this and I just can't find a solution; I've tried finding the derivative of $(f(x))^2$ and evaluating that derivative at $0$, but no matter how I attack it I always get an answer of $0$ when the derivative is evaluated at $0$. Any help would be appreciated!
Best Answer
First, when $x = 0$, $y = (f(0))^2 = 9$. Then,
\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}}{\mathrm{d}x} (f(x))^2 \\ &= 2 f(x) \frac{\mathrm{d}}{\mathrm{d}x} f(x) \\ &= 2 f(x) f'(x) \end{align*}
So \begin{align*} \left. \frac{\mathrm{d}y}{\mathrm{d}x} \right|_{x = 0} &= 2 f(0) f'(0) \\ &= 2 \cdot 3 \cdot (\cos(\pi \cdot 0) + 0^4+6) \\ &= 6 \cdot 7 \\ &= 42 \text{.} \end{align*}
Then an equation of the line is $y - 9 = 42(x-0)$.