If y = fleft( frac{5x + 1}{10x^2 - 3} right) | vedclass

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(A) Given $y = f\left( \frac{5x + 1}{10x^2 - 3} \right)$ and $f'(x) = \cos x$. Let $u = \frac{5x + 1}{10x^2 - 3}$. Then $y = f(u)$. By the chain rule,

Vedclass · Accepted Answer

$\cos \left( \frac{5x + 1}{10x^2 - 3} \right) \frac{d}{dx} \left( \frac{5x + 1}{10x^2 - 3} \right)$

Vedclass · Answer

(A) Given $y = f\left( \frac{5x + 1}{10x^2 - 3} \right)$ and $f'(x) = \cos x$. Let $u = \frac{5x + 1}{10x^2 - 3}$. Then $y = f(u)$. By the chain rule,$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. Since $y = f(u)$,we have $\frac{dy}{du} = f'(u) = \cos(u)$. Substituting $u$ back,we get $\frac{dy}{du} = \cos \left( \frac{5x + 1}{10x^2 - 3} \right)$. Therefore,$\frac{dy}{dx} = \cos \left( \frac{5x + 1}{10x^2 - 3} \right) \cdot \frac{d}{dx} \left( \frac{5x + 1}{10x^2 - 3} \right)$.

If $y = f\left( \frac{5x + 1}{10x^2 - 3} \right)$ and $f'(x) = \cos x$,then $\frac{dy}{dx} = $

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