If y = e^{4x} left( frac{x-4}{x+3} right)^{fr | vedclass

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(A) Given $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$.Taking natural logarithm on both sides:$\ln y = \ln \left( e^{4x} \left( \frac{x-4

Vedclass · Accepted Answer

$\frac{dy}{dx} = y \left[ 4 + \frac{21}{4(x-4)(x+3)} \right]$

Vedclass · Answer

(A) Given $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$.Taking natural logarithm on both sides:$\ln y = \ln \left( e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}} \right)$$\ln y = \ln(e^{4x}) + \ln \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$$\ln y = 4x + \frac{3}{4} [\ln(x-4) - \ln(x+3)]$Differentiating with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{1}{x-4} - \frac{1}{x+3} \right)$$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{(x+3) - (x-4)}{(x-4)(x+3)} \right)$$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{3}{4} \left( \frac{7}{(x-4)(x+3)} \right)$$\frac{1}{y} \frac{dy}{dx} = 4 + \frac{21}{4(x-4)(x+3)}$Therefore,$\frac{dy}{dx} = y \left[ 4 + \frac{21}{4(x-4)(x+3)} \right]$.

If $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$,then $\frac{dy}{dx} = $

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