## Calculus: Early Transcendentals 8th Edition

$f'(1)=\frac{-16}{13}$
Take the derivative as is on either side of the equation: $f'(x)+3x^2[f(x)]^2f'(x)+2x[f(x)]^3=0$ Move all terms with f'(x) onto one side of the equal sign and distribute the f'(x) out of each term: $f'(x)(1+3x^2[f(x)]^2)=-2x[f(x)]^3$ Isolate f'(x) by dividing both sides by the terms: $f'(x)=\frac{-2x[f(x)]^3}{1+3x^2[f(x)]^2}$ Plug in 1 to find f'(1): $f'(1)=\frac{-2(1)[f(1)]^3}{1+3(1)^2[f(1)]^2}$ $f'(1)=\frac{-2(8)}{1+3(4)}=\frac{-16}{13}$