How do you differentiate f(x)=(x+2)^2(x-5)^3 using the product rule?

Feb 6, 2016

$f ' \left(x\right) = {\left(x - 5\right)}^{2} \left(x + 2\right) \left(5 x - 4\right)$

Explanation:

for a function f)x) = g(x).h(x) ie. a product of 2 functions

then f'(x) = g(x).h'(x) + h(x).g'(x).............................(A)

$\textcolor{b l a c k}{\text{-------------------------------------}}$
here g(x) $= {\left(x + 2\right)}^{2}$

and using$\textcolor{b l u e}{\text{ chain rule }}$

g'(x) $= 2 \left(x + 2\right) \frac{d}{\mathrm{dx}} \left(x + 2\right) = 2 \left(x + 2\right) .1 = 2 \left(x + 2\right)$
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and h(x) $= {\left(x - 5\right)}^{3}$
$\textcolor{b l u e}{\text{ again using chain rule }}$

h'(x)$= 3 {\left(x - 5\right)}^{2} \frac{d}{\mathrm{dx}} \left(x - 5\right) = 3 {\left(x - 5\right)}^{2} .1 = 3 {\left(x - 5\right)}^{2}$
$\textcolor{b l a c k}{\text{--------------------------------------------------}}$

substituting back into ( A) gives :

f'(x) $= {\left(x + 2\right)}^{2} . 3 {\left(x - 5\right)}^{2} + {\left(x - 5\right)}^{3} .2 \left(x + 2\right)$

$= 3 {\left(x + 2\right)}^{2} {\left(x - 5\right)}^{2} + {\left(x - 5\right)}^{3.} 2 \left(x + 2\right)$

take out common factors $\left(x + 2\right) {\left(x - 5\right)}^{2}$

$= {\left(x - 5\right)}^{2} \left(x + 2\right) \left[3 \left(x + 2\right) + 2 \left(x - 5\right)\right]$

$\Rightarrow f ' \left(x\right) = {\left(x - 5\right)}^{2} \left(x + 2\right) \left[3 x + 6 + 2 x - 10\right] = {\left(x - 5\right)}^{2} \left(x + 2\right) \left(5 x - 4\right)$

$\textcolor{b l a c k}{\text{----------------------------------------------------}}$