Derivative

[kickingtoad]

I am tripping on finding the derivative of this.

$\displaystyle {f(x)=(-3x^{2}-3)^{7}(2x^{2}-7)^{14}}$

find f'(x)

$\displaystyle {f'(x)=(F'(x)G(x))G'(x)}$

This is what I've been doing...

1. $\displaystyle {7(-3x^{2}-3)^{6}(-6x)14(2x^{2}-7)^{13}(4x}...$

How do I know which part to use for $\displaystyle G'(x)$ at the end?

 

[DrSteve]

This function is a product of two functions, so you must use the product rule. Each factor in the product is a composition of functions, so you will apply the chain rule twice, as you are doing the product rule. Try it again and I'll help you more if you're stuck.

 

[kickingtoad]

\(\displaystyle {7(-3x^{2}-3)^{6}(-3x^{2}-3)(-6x)+14(2x^{2}-7)^{13}(2x^{2}-7)(4x)\)

What to do from here?

 

[DrSteve]

Your answer is almost correct. You added an extra factor of

$\displaystyle -3x^2-3$

in the first term, and an extra factor of

$\displaystyle 2x^2-7$

in the last term. Just get rid of these two factors and the answer is right.

 

[soroban]

Hello, kickingtoad!

Differentiate: .\(\displaystyle h(x) \:=\-3x^{2}-3)^7(2x^{2}-7)^{14}\)

I did it like this . . .

$\displaystyle h(x) \;=\;\left[-3(x^2+1)\right]^7(2x^2-7)^{14} \;=\; (-3)^7(x^2+1)^7(2x^2-7)^{14} \;=\;-2187\overbrace{(x^2+1)^7}^{f(x)}\overbrace{(2x^2-7)^{14}}^{g(x)}$


$\displaystyle \text{Product Rule: }\;h'(x) \;=\;-2187\cdot \left[\overbrace{(x^2+1)^7}^{f(x)} \cdot \overbrace{14(2x^2-7)^{13}\cdot 4x}^{g'(x)} \;+\; \overbrace{7(x^2+1)^6\cdot2x}^{f '(x)} \cdot \overbrace{(2x^2-7)^{14}}^{g(x)}\right]$

. . . . $\displaystyle h'(x) \;=\;-2187\bigg[56x(x^2+1)^7(2x^2-7)^{13} + 14x(x^2+1)^6(2x^2-7)^{14}\bigg]$

. . . . . . . . $\displaystyle =\;-2187(14x)(x^2+1)^6(2x^2-7)^{13}\bigg[4(x^2+1) + (2x^2-7)\bigg]$

. . . . . . . . $\displaystyle =\;-30,\!618\:\!x(x^2+1)^6(2x^2-7)^{13}\left[4x^2 + 4 + 2x^2 - 7\right]$

. . . . . . . . $\displaystyle =\;-30,\!618\:\!x(x^2+1)^6(2x^2-7)^{13}(6x^2 - 3)$

. . . . . . . . $\displaystyle =\;-91,\!854\:\!x(x^2+1)^6(2x^2-7)^{13}(2x^2-1)$

 

[lookagain]

I am tripping on finding the derivative of this.

$\displaystyle {f(x)=(-3x^{2}-3)^{7}(2x^{2}-7)^{14}}$

find f'(x)

kickingtoad,

I would suggest you consider factoring out a $\displaystyle -3$ first:

$\displaystyle f(x) = [-3(x^2 - 1)]^7(2x^2 - 7)^{14} =$

$\displaystyle (-3)^7(x^2 - 1)^7(2x^2 - 7)^{14} =$

$\displaystyle -2,187(x^2 - 1)^7(2x^2 - 7)^{14}$

Then concentrate on the product rule for the product of the last two factors,
which are the polynomial factors, as you carry down $\displaystyle -2,187$ as a
multiplier on that part of the evolving simplification of the derivative (result).