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Annette [7]
3 years ago
15

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

Mathematics
1 answer:
il63 [147K]3 years ago
3 0

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}}

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Step-by-step explanation:

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