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Continuity and Differentiability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSContinuity and Differentiability3 Marks
Question
Differentiate the function $(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4} w.r.t. x.$
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Answer

Given: $(x + 3)^2.(x + 4)^3.(x + 5)^4$
Let $y = (x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$
Taking $\log$ on both sides, we get
$\log y = \log \left((x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}\right)$
$\Rightarrow \log y=\log (x+3)^{2}+\log (x+4)^{3}+\log (x+5)^{4}$
$\Rightarrow \log y=2 \cdot \log (x+3)+3 \cdot \log (x+4)+4 \cdot \log (x+5)$
Now, differentiating both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{y})=\frac{\mathrm{d}}{\mathrm{dx}}(2 \cdot \log (\mathrm{x}+3))+\frac{\mathrm{d}}{\mathrm{dx}}(3 \cdot \log (\mathrm{x}+4))+\frac{\mathrm{d}}{\mathrm{dx}}(4 \cdot \log (\mathrm{x}+5))$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=2 \cdot \frac{1}{x+3} \cdot \frac{d}{d x}(x+3)+3 \cdot \frac{1}{x+4} \cdot \frac{d}{d x}(x+4)+4 \cdot \frac{1}{x+5} \cdot \frac{d}{d x}(x+5)$
$\Rightarrow \frac{d y}{d x}=y\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]$
$\Rightarrow \frac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^4$ $\left[\frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\right]$
$ \Rightarrow \frac{\mathrm{d} y}{\mathrm{dx}}=(\mathrm{x}+ 3)^{1}(\mathrm{x}+4)^{2}(\mathrm{x}+5)^{3}\left[2\left(\mathrm{x}^{2}+9 \mathrm{x}+20\right)+3\left(\mathrm{x}^{2}+8 \mathrm{x}+15\right)\right. \left.+4\left(\mathrm{x}^{2}+7 \mathrm{x}+12\right)\right] $
$=$ $(x+3)(x+4)^{2}(x+5)^{3}\left(9 x^{2}+70 x+133\right)$

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