Differentiate the function with respect to x.

Question:

Differentiate the function with respect to x.

$(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$

Solution:

Let $y=(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$

Taking logarithm on both the sides, we obtain

$\log y=\log (x+3)^{2}+\log (x+4)^{3}+\log (x+5)^{4}$

$\Rightarrow \log y=2 \log (x+3)+3 \log (x+4)+4 \log (x+5)$

Differentiating both sides with respect to x, we obtain

$\frac{1}{y} \cdot \frac{d y}{d x}=2 \cdot \frac{1}{x+3} \cdot \frac{d}{d x}(x+3)+3 \cdot \frac{1}{x+4} \cdot \frac{d}{d x}(x+4)+4 \cdot \frac{1}{x+5} \cdot \frac{d}{d x}(x+5)$

$\Rightarrow \frac{d y}{d x}=y\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]$

$\Rightarrow \frac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^{4} \cdot\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]$

$\Rightarrow \frac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^{4} \cdot\left[\frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\right]$

$\Rightarrow \frac{d y}{d x}=(x+3)(x+4)^{2}(x+5)^{3} \cdot\left[2\left(x^{2}+9 x+20\right)+3\left(x^{2}+8 x+15\right)+4\left(x^{2}+7 x+12\right)\right]$

$\therefore \frac{d y}{d x}=(x+3)(x+4)^{2}(x+5)^{3}\left(9 x^{2}+70 x+133\right)$