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Differentiate the functions given in w.r.t. $x : $ $ (x + 3)^2 \: . \: (x + 4)^3 \: . \: (x + 5)^4 $

$\begin{array}{1 1}(x+3)^2.(x+4)^3.(x+5)^4.\begin{bmatrix}\large\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\end{bmatrix} \\ (x+3)^2.(x+4)^3.(x+5)^4.\begin{bmatrix}\large\frac{2}{x-3}+\frac{3}{x-4}+\frac{4}{x+5}\end{bmatrix} \\ (x+3)^2.(x+4)^3.\begin{bmatrix}\large\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\end{bmatrix} \\ (x+3)^2.(x+4)^3.(x+5)^4.\begin{bmatrix}\large\frac{4}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\end{bmatrix}\end{array} $

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Toolbox:
  • $\log mn=\log m+\log n$
  • $\log m^{\large n}=n\log m$
Step 1:
Let $(x+3)^2.(x+4)^3.(x+5)^4$
Taking $\log$ on both sides
$\log y=\log[(x+3)^2.(x+4)^3.(x+5)^4]$
We know that $\log mn=\log m+\log n$
$\Rightarrow \log(x+3)^2+\log(x+4)^3+\log(x+5)^4$
$\Rightarrow 2\log(x+3)+3\log(x+4)+4\log(x+5)$
Step 2:
Differentiating both sides with respect to $x$
$\large\frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}$
$\large\frac{dy}{dx}$$=y.\begin{bmatrix}\large\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\end{bmatrix}$
By substituting the value of $y$ in the above equation.
$\large\frac{dy}{dx}$$=(x+3)^2.(x+4)^3.(x+5)^4.\begin{bmatrix}\large\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\end{bmatrix}$

 

answered May 8, 2013 by sreemathi.v
edited Jun 16, 2013 by vijayalakshmi_ramakrishnans
 

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