Expert answers to "If f(x)=x2[f(x)]4=18 and f(1)=2, find f′(1)."

oliviayychengwh

Answered question

2021-12-16

f (x) = x^{2} {[f (x)]}^{4} = 18

and

f (1) = 2

, find

f^{'} (1)

Mary Goodson

Beginner2021-12-17Added 37 answers

f (x) + x^{2} {[f (x)]}^{4} = 18

differentiating on both sides

\Rightarrow f^{'} (x) + {(2 x) [f (x)] 4 + x 24 [f (x)] 3 f^{'} (x)} = 0

since

{(U V)}^{'} = U^{'} V + U V^{'}; \frac{d}{d x} x^{n} = n x^{n - 1}

\Rightarrow f^{'} (x) + (2 x) {[f (x)]}^{4} + x^{2} 4 {[f (x)]}^{3} f^{'} (x) = 0

\Rightarrow f^{'} (x) + x^{2} 4 {[f (x)]}^{3} f^{'} (x) + (2 x) {[f (x)]}^{4} = 0

\Rightarrow f^{'} (x) + x^{2} 4 {[f (x)]}^{3} f^{'} (x) = - (2 x) {[f (x)]}^{4}

\Rightarrow {1 + x^{2} 4 {[f (x)]}^{3}} f^{'} (x) = - (2 x) {[f (x)]}^{4}

\Rightarrow f^{'} (x) = - (2 x) \frac{{[f (x)]}^{4}}{1 + x^{2} 4 {[f (x)]}^{3}}

\Rightarrow f^{'} (1) = - (2 (1)) \frac{{[f (1)]}^{4}}{1 + {(1)}^{2} 4 {[f (1)]}^{3}}

\Rightarrow f^{'} (1) = - (2) \frac{{[2]}^{4}}{1 + 4 {[2]}^{3}}

\Rightarrow f^{'} (1) = - \frac{32}{33}

turtletalk75

Beginner2021-12-18Added 29 answers

Shortly answer?

nick1337

Expert2021-12-28Added 777 answers

$f (x) + x^{2} [f (x)]^{4} = 18$

$\Rightarrow f^{'} (1) = - \frac{32}{33}$

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