## Calculus 10th Edition

$y'=-12\sec^3{4x}\tan{4x}.$ The derivative evaluated at $(0, 25)$ is $0.$
$y=f(x)+g(x)\rightarrow f(x)=26$; $g(x)=-\sec^3{4x}$ $f'(x)=0$ $g'(x)$ is found using the Chain Rule: $u=\sec{4x}$; $\dfrac{du}{dx}=4\sec{4x}\tan{4x}$ $y(u)=-u^3;\dfrac{d}{du}g(u)=-3u^2$ $\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=-12\sec^3{4x}\tan{4x}.$ $y'=f'(x)+g'(x)=0-12\sec^3{4x}\tan{4x}=-12\sec^3{4x}\tan{4x}.$ To evaluate the derivative, plug in $x=0\rightarrow -12\sec^3{0}\tan{0}=0.$ A graphing utility was used to verify this result.