If y = f left( frac{3x + 4}{5x + 6} right) an | vedclass

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(B) Given $y = f \left( \frac{3x + 4}{5x + 6} ight)$ and $f'(x) = 	an(x^2)$.Applying the chain rule to differentiate with respect to $x$:$\frac{dy}

Vedclass · Accepted Answer

$- 2 	an \left[ \frac{3x + 4}{5x + 6} ight]^2 \cdot \frac{1}{(5x + 6)^2}$

Vedclass · Answer

(B) Given $y = f \left( \frac{3x + 4}{5x + 6} ight)$ and $f'(x) = 	an(x^2)$.Applying the chain rule to differentiate with respect to $x$:$\frac{dy}{dx} = f' \left( \frac{3x + 4}{5x + 6} ight) \cdot \frac{d}{dx} \left( \frac{3x + 4}{5x + 6} ight)$Using the quotient rule for the derivative of the inner function:$\frac{d}{dx} \left( \frac{3x + 4}{5x + 6} ight) = \frac{3(5x + 6) - 5(3x + 4)}{(5x + 6)^2} = \frac{15x + 18 - 15x - 20}{(5x + 6)^2} = \frac{-2}{(5x + 6)^2}$Substituting $f'(x) = 	an(x^2)$ into the expression:$\frac{dy}{dx} = 	an \left( \frac{3x + 4}{5x + 6} ight)^2 \cdot \frac{-2}{(5x + 6)^2}$Thus,the correct option is $B$.

If $y = f \left( \frac{3x + 4}{5x + 6} \right)$ and $f'(x) = \tan(x^2)$,then $\frac{dy}{dx} = $

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