If f(x^{5}) = 5x^{3},then f'(x) is equal to | vedclass

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(A) Given,$f(x^{5}) = 5x^{3}$.Let $x^{5} = y$,which implies $x = y^{1/5}$.Then $x^{3} = (y^{1/5})^{3} = y^{3/5}$.Substituting this into the function,w

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$\frac{3}{\sqrt[5]{x^{2}}}$

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(A) Given,$f(x^{5}) = 5x^{3}$.Let $x^{5} = y$,which implies $x = y^{1/5}$.Then $x^{3} = (y^{1/5})^{3} = y^{3/5}$.Substituting this into the function,we get $f(y) = 5y^{3/5}$.Replacing $y$ with $x$,we have $f(x) = 5x^{3/5}$.Now,differentiating $f(x)$ with respect to $x$:$f'(x) = 5 \cdot \frac{3}{5} x^{(3/5 - 1)} = 3x^{-2/5}$.This can be rewritten as $f'(x) = \frac{3}{x^{2/5}} = \frac{3}{\sqrt[5]{x^{2}}}$.

If $f(x^{5}) = 5x^{3}$,then $f'(x)$ is equal to

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