Find d y d x , if y = (3 x-4)^3 (x+1)^4(x+2) - Vidyadip

Question

Find $\frac{d y}{d x}$, if $y =\sqrt{\frac{(3 x-4)^3}{(x+1)^4(x+2)}}$

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Answer

$
\begin{aligned}
& y=\sqrt{\frac{(3 x-4)^3}{(x+1)^4(x+2)}} \\
& =\frac{(3 x-4)^{\frac{3}{2}}}{(x+1)^{\frac{4}{2}} \cdot(x+2)^{\frac{1}{2}}}
\end{aligned}
$
Taking logarithm of both sides, we get
$
\begin{aligned}
& \log y=\log \left[\frac{(3 x-4)^{\frac{3}{2}}}{(x+1)^{\frac{4}{2}} \cdot(x+2)^{\frac{1}{2}}}\right] \\
& =\log (3 x-4)^{\frac{3}{2}}-\left[\log (x+1)^2+\log (x+2)^{\frac{1}{2}}\right] \\
& =\frac{3}{2} \log (3 x-4)-2 \log (x+1)-\frac{1}{2} \log (x+2)
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& \frac{1}{y} \cdot \frac{d y}{d x}=\frac{3}{2} \cdot \frac{d}{d x}[\log (3 x-4)]-2 \frac{d}{d x}[\log (x+1)]-\frac{1}{2} \cdot \frac{d}{d x}[\log (x+2)] \\
& =\frac{3}{2} \cdot \frac{1}{3 x-4} \cdot \frac{d}{d x}(3 x-4)-2 \cdot \frac{1}{x+1} \cdot \frac{d}{d x}(x+1)-\frac{1}{2} \cdot \frac{1}{x+2} \cdot \frac{d}{d x}(x+2) \\
& \therefore \frac{1}{y} \cdot \frac{d y}{d x}=\frac{3}{2(3 x-4)} \times 3-\frac{2}{x+1} \times 1-\frac{1}{2(x+2)} \times 1 \\
& \therefore \frac{1}{y} \cdot \frac{d y}{d x}=\frac{9}{2(3 x-4)}-\frac{2}{x+1}-\frac{1}{2(x+2)} \\
& \therefore \frac{d y}{d x}=\frac{y}{2}\left[\frac{9}{3 x-4}-\frac{4}{x+1}-\frac{1}{x+2}\right] \\
& \therefore \frac{d y}{d x}=\frac{1}{2} \sqrt{\frac{(3 x-4)^3}{(x+1)^4(x+2)}}\left[\frac{9}{3 x-4}-\frac{4}{x+1}-\frac{1}{x+2}\right]
\end{aligned}
$

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