Derivative, Derivative, Derivative

Calculus Level 2

If $f(x)=(x+1)^3(x^2-1)^2$ , what is the value of $f'(2)$ ?

Details and assumptions

$f'(x)$ denotes the derivative of $f(x)$ .

The answer is 891.

4 solutions

Hobart Pao
Feb 4, 2015

I used chain rule on (x+1)^3 and (x^2 - 1)^2 separately. Then I applied product rule.

Vikram Waradpande
Mar 5, 2014

Let $y=f(x)=(x+1)^3*(x^2-1)^2$ Take $log$ of both sides and differentiate with respect to $x$ to get

$\large{ \frac{dy}{dx}} = (x+1)^3*(x^2-1)^2 \left ( \frac{3}{x+1} + \frac{4x}{x^2-1} \right )$ Now put $x=2$ to get $f'(2) = 891$

Don't you think you've to make sure that $f(x)$ attains positive real values before taking $\log$ on both sides ( assuming $f(x)$ to be real valued ) ?

Notice that, $f(x)\,>\,0\,\,\forall \,\,x\,\,\in \,\, (-1,1)\cup (1,\infty)$ , so, in this interval $\f^{'}(x)\,=\,f(x) \cdot \Bigg( \dfrac{3}{x+1} + \dfrac{4x}{x^{2}-1} )$ .

Note that, $2\,\,\in\,\,(-1,-1)\,\cup\,(1,\infty)$ , hence, we can substitute $x=2$ in the obtained expression of $f^{'}(x)$ to get $f^{'}(2)\,=\,891$ .

Aditya Sky - 4 years, 8 months ago

Vikas Vishwakarma
Mar 5, 2014

it is little bit long, but you will get 891 surely

Andrew Tiu
Jan 9, 2014

Just expand the function: f(x)=x^7+3x^6+x^5-5x^4-5x^3+x^2+3x+1--->Now find derivative using the all mighty power rule: f'(x)7x^6+18x^5+5x^4-20x^3-15x^2+2x+3---> Substitute 2 for x (I'm skipping the cumbersome algebra as I know you all can understand). Answer is 891

Derivative, Derivative, Derivative

The answer is 891.

4 solutions

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