If y = frac{a + b{x^{3/2}}}{{x^{5/4}}} and y' | vedclass

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Question

(A) Given $y = \frac{a + b{x^{3/2}}}{{x^{5/4}}} = a{x^{-5/4}} + b{x^{1/4}}$.Differentiating with respect to $x$:$y' = a(-\frac{5}{4}){x^{-9/4}} + b(\f

Vedclass · Accepted Answer

$\sqrt{5} : 1$

Vedclass · Answer

(A) Given $y = \frac{a + b{x^{3/2}}}{{x^{5/4}}} = a{x^{-5/4}} + b{x^{1/4}}$.Differentiating with respect to $x$:$y' = a(-\frac{5}{4}){x^{-9/4}} + b(\frac{1}{4}){x^{-3/4}}$.Setting $y' = 0$ at $x = 5$:$-\frac{5a}{4}{x^{-9/4}} + \frac{b}{4}{x^{-3/4}} = 0$.Multiply by $4{x^{9/4}}$:$-5a + b{x^{6/4}} = 0$.$-5a + b{x^{3/2}} = 0$.At $x = 5$,we have $b(5)^{3/2} = 5a$.$\frac{a}{b} = \frac{5^{3/2}}{5} = 5^{3/2 - 1} = 5^{1/2} = \sqrt{5}$.Thus,the ratio $a:b = \sqrt{5} : 1$.

If $y = \frac{a + b{x^{3/2}}}{{x^{5/4}}}$ and $y' = 0$ at $x = 5$,then the ratio $a:b$ is equal to

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