Question

Differentiate the given function with respect to x.
${{\left( x+3 \right)}^{2}}\times {{\left( x+4 \right)}^{3}}\times {{\left( x+5 \right)}^{4}}$

Answer 1

Hint: As the expression contains multiplication of few algebraic expressions, use logarithm to separate them and then differentiate them separately.

Complete step-by-step answer:
Let us assume the function to be y.
So by assuming, we get:
$y={{\left( x+3 \right)}^{2}}\times {{\left( x+4 \right)}^{3}}\times {{\left( x+5 \right)}^{4}}$ …..(1)
By taking log on both sides, we get:
\log y=\log \left\\{ {{\left( x+3 \right)}^{2}}\times {{\left( x+4 \right)}^{3}}\times {{\left( x+5 \right)}^{4}} \right\\}
By applying logarithm property,
log(a.b)=log(a)+log(b) …..(2),we get:
\log y=\log \left\\{ {{\left( x+3 \right)}^{2}} \right\\}+\log \left\\{ {{\left( x+4 \right)}^{3}} \right\\}+\log \left\\{ {{\left( x+5 \right)}^{4}} \right\\}
By applying logarithm property,
$\log \left( {{a}^{b}} \right)=b\log \left( a \right)$ …..(3), we get:
\log y=2\log \left\\{ \left( x+3 \right) \right\\}+3\log \left\\{ \left( x+4 \right) \right\\}+4\log \left\\{ \left( x+5 \right) \right\\}
By differentiating both sides w.r.t. x, we get:
$\dfrac{d\left( \log y \right)}{dx}=\dfrac{d\left( 2\log \left( x+3 \right)+3\log \left( x+4 \right)+4\log \left( x+5 \right) \right)}{dx}$
By applying differentiation properties,
$\dfrac{d(a+b)}{dx}=\dfrac{da}{dx}+\dfrac{db}{dx}$ , we get:
$\dfrac{d\left( \log y \right)}{dx}=\dfrac{d\left( 2\log \left( x+3 \right) \right)}{dx}+\dfrac{d\left( 3\log \left( x+4 \right) \right)}{dx}+\dfrac{d\left( 4\log \left( x+5 \right) \right)}{dx}$
By applying differentiation properties,
$\dfrac{d\left( k\times f\left( x \right) \right)}{dx}=k\times \dfrac{d\left( f\left( x \right) \right)}{dx}$
In the above equation, f(x) is a function of x and k is a constant.
By above, we get:
$\dfrac{d\left( \log y \right)}{dx}=2\dfrac{d\left( \log \left( x+3 \right) \right)}{dx}+3\dfrac{d\left( \log \left( x+4 \right) \right)}{dx}+4\dfrac{d\left( \log \left( x+5 \right) \right)}{dx}$
By applying differentiation properties,
$\dfrac{d\left( \log\left (f\left( x \right)\right) \right)}{dx}=\dfrac{1}{f\left( x \right)}\times \dfrac{d\left( f\left( x \right) \right)}{dx}$
In the above equation, f(x) is a function of x.
By above, we get:
$\dfrac{1}{y}\dfrac{d\left( y \right)}{dx}=2\dfrac{1}{\left( x+3 \right)}\dfrac{d\left( x+3\right)}{dx}+3\dfrac{1}{\left( x+4 \right)}\dfrac{d\left( x+4 \right)}{dx}+4\dfrac{1}{\left( x+5 \right)}\dfrac{d\left( x+5 \right)}{dx}$
By using equation (2), we get:
$\dfrac{1}{y}\dfrac{d\left( y \right)}{dx}=2\dfrac{1}{\left( x+3 \right)}\left( \dfrac{dx}{dx}+\dfrac{d\left( 3 \right)}{dx} \right)+3\dfrac{1}{\left( x+4 \right)}\left( \dfrac{dx}{dx}+\dfrac{d\left( 4 \right)}{dx} \right)+4\dfrac{1}{\left( x+5 \right)}\left( \dfrac{dx}{dx}+\dfrac{d\left( 5 \right)}{dx} \right)$
We know differentiation of a constant with respect to x is 0 and differentiation of x with respect to x is 1.
By using these, we get:

& \dfrac{1}{y}\dfrac{d\left( y \right)}{dx}=2\dfrac{1}{\left( x+3 \right)}\left( 1+0 \right)+3\dfrac{1}{\left( x+4 \right)}\left( 1+0 \right)+4\dfrac{1}{\left( x+5 \right)}\left( 1+0 \right) \\\ & \dfrac{1}{y}\dfrac{d\left( y \right)}{dx}=\dfrac{2}{\left( x+3 \right)}+\dfrac{3}{\left( x+4 \right)}+\dfrac{4}{\left( x+5 \right)} \\\ \end{aligned}$$ So by multiplying y on both sides, we get: $$\dfrac{d\left( y \right)}{dx}=y\left( \dfrac{2}{\left( x+3 \right)}+\dfrac{3}{\left( x+4 \right)}+\dfrac{4}{\left( x+5 \right)} \right)$$ By substituting the value of y, we get: $$\dfrac{dy}{dx}=\left( {{\left( x+3 \right)}^{2}}\times {{\left( x+4 \right)}^{3}}\times {{\left( x+5 \right)}^{4}} \right)\left( \dfrac{2}{\left( x+3 \right)}+\dfrac{3}{\left( x+4 \right)}+\dfrac{4}{\left( x+5 \right)} \right)$$ By simplifying, we get: $$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{{{\left( x+3 \right)}^{2}}{{\left( x+4 \right)}^{3}}{{\left( x+5 \right)}^{4}}}{\left( x+3 \right)\left( x+4 \right)\left( x+5 \right)}\left( 2\left( {{x}^{2}}+4x+5x+20 \right)+3\left( {{x}^{2}}+3x+5x+15 \right)+4\left( {{x}^{2}}+3x+4x+12 \right) \right) \\\ & \dfrac{dy}{dx}=\dfrac{\left( x+3 \right){{\left( x+4 \right)}^{2}}{{\left( x+5 \right)}^{3}}}{1}\left( 2\left( {{x}^{2}}+9x+20 \right)+3\left( {{x}^{2}}+8x+15 \right)+4\left( {{x}^{2}}+7x+12 \right) \right) \\\ & \\\ \end{aligned}$$By solving more and cancelling terms, we get: $$\begin{aligned} & \dfrac{dy}{dx}=\left( x+3 \right){{\left( x+4 \right)}^{2}}{{\left( x+5 \right)}^{3}}\left( \left( 2{{x}^{2}}+18x+40 \right)+\left( 3{{x}^{2}}+24x+45 \right)+\left( 4{{x}^{2}}+28x+48 \right) \right) \\\ & \dfrac{dy}{dx}=\left( x+3 \right){{\left( x+4 \right)}^{2}}{{\left( x+5 \right)}^{3}}\left( 9{{x}^{2}}+70x+133 \right) \\\ \end{aligned}$$ So the differentiation of given function is as follows: $$\dfrac{d\left( {{\left( x+3 \right)}^{2}}\times {{\left( x+4 \right)}^{3}}\times {{\left( x+5 \right)}^{4}} \right)}{dx}=\left( x+3 \right){{\left( x+4 \right)}^{2}}{{\left( x+5 \right)}^{3}}\left( 9{{x}^{2}}+70x+133 \right)$$ Note: Now let’s write two equations,observe carefully and don’t get confused in these aspects. $$\begin{aligned} & \log \left( a\times b \right)=\log a+\log b\text{ ,but} \\\ & \dfrac{d\left( a\times b \right)}{dx}\ne \dfrac{da}{dx}+\dfrac{db}{dx} \\\ \end{aligned}$$ So whenever you see multiplication. First apply logarithm and then differentiate.