A function f,defined for all positive real nu | vedclass

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(B) Given the equation $f(x^2) = x^3$ for $x > 0$.To find $f'(4)$,we differentiate both sides of the equation with respect to $x$ using the chain rule

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given the equation $f(x^2) = x^3$ for $x > 0$.To find $f'(4)$,we differentiate both sides of the equation with respect to $x$ using the chain rule:$\frac{d}{dx}[f(x^2)] = \frac{d}{dx}[x^3]$$f'(x^2) \cdot (2x) = 3x^2$For $x > 0$,we can divide by $2x$:$f'(x^2) = \frac{3x^2}{2x} = \frac{3}{2}x$We want to find $f'(4)$. Since $x^2 = 4$ and $x > 0$,we have $x = 2$.Substituting $x = 2$ into the expression for $f'(x^2)$:$f'(4) = \frac{3}{2}(2) = 3$Thus,the value of $f'(4)$ is $3$.

$A$ function $f$,defined for all positive real numbers,satisfies the equation $f(x^2) = x^3$ for every $x > 0$. Then the value of $f'(4) =$

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