Question

Find f'(1) if 6x*(f(x))^4 + 9x^3 f(x) = 78 and f(1) = -2​

Answer 1

Differentiate both sides of

$6x f(x)^4 + 9x^3 f(x) = 78$

with respect to x :

$\left(6x f(x)^4 + 9x^3 f(x)\right)' = (78)' \\\\ 6\left(xf(x)^4\right)' + 9\left(x^3f(x)\right)' = 0$

Use the product and chain rules for the left side:

$\left(xf(x)^4\right)' = (x)'f(x)^4 + x\left(f(x)^4\right)' = f(x)^4 + 4xf(x)^3f'(x)$

$\left(x^3f(x)\right)' = \left(x^3\right)'f(x) + x^3f'(x) = 3x^2f(x)+x^3f'(x)$

Solve for f '(x) :

$6\left(f(x)^4+4xf(x)^3f'(x)\right) + 9\left(3x^2f(x)+x^3f'(x)\right) = 0 \\\\ 6f(x)^4+27x^2f(x) + \left(24xf(x)^3+9x^3\right)f'(x) = 0 \\\\ f'(x) = -\dfrac{6f(x)^4+27x^2f(x)}{24xf(x)^3+9x^3}$

Then

$f'(1) = \boxed{\dfrac{14}{61}}$