Answer
$y=\dfrac{8x-7}{5}.$
Work Step by Step
Using the Chain Rule:
$u=2x^2-7$; $\dfrac{du}{dx}=4x$
$\dfrac{d}{du}f(u)=\dfrac{1}{2\sqrt{u}}$
$\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=\dfrac{2x}{\sqrt{2x^2-7}}.$
$f'(4)=\dfrac{2(4)}{\sqrt{2(4)^2-7}}=\dfrac{8}{5}.$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-5)=\frac{8}{5}(x-4)\rightarrow y=\dfrac{8x-7}{5}.$
A graphing calculator and a computer algebra system have been used to confirm these results.
