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Differentiate the following w.r.t $x$: $(x+1)^2(x+2)^3(x+3)^4$.

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1 Answer

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Toolbox:
  • If $y=[f(x)]^{\large g(x)}$.
  • Taking $ \log$ on both sides we get,
  • We have $\log y=g(x).\log[f(x)]$
Step 1:
$y=(x+1)^2(x+2)^3(x+3)^4$
Take $log$ on both sides,
We know $\log(abc)=\log a+\log b+\log c$
$\log y=\log(x+1)^2+\log(x+2)^3+\log(x+3)^4$
Also $\log x^a=a\log x$
$\log y=2\log(x+1)+3\log(x+2)+4\log(x+3)$
Now differentiating on both sides,
$\large\frac{1}{y}\frac{dy}{dx}$$=2.\large\frac{1}{x+1}$$+3.\large\frac{1}{x+2}$$+4.\large\frac{1}{x+3}$
Therefore $\large\frac{dy}{dx}=$$y\bigg[\large\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}\bigg]$
$\qquad\qquad\;\;\;\;=(x+1)^2(x+2)^3(x+3)^4\bigg[\large\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}\bigg]$
Step 2:
On simplifying we get,
$\qquad\qquad\;\;\;\;=(x+1)(x+2)^2(x+3)^3[2(x+2)(x+3)+3(x+1)(x+3)+4(x+1)(x+3)]$
$\qquad\qquad\;\;\;\;=(x+1)(x+2)^2(x+3)^3[2(x^2+5x+6)+3(x^2+4x+3)+4(x^2+3x+2)]$
$\qquad\qquad\;\;\;\;=(x+1)(x+2)^2(x+3)^3[2x^2+10x+12+3x^2+12x+9+4x^2+12x+8]$
$\qquad\qquad\;\;\;\;=(x+1)(x+2)^2(x+3)^3[9x^2+34x+29]$
answered Jun 27, 2013 by sreemathi.v
 
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