Question

Let $f\left(x\right)=x^5+{2x}^3+3x+4$ then find the value of $28\frac{d}{dx}\left(f^{{}^{\prime }}\right(x\left)\right)$ at $x=-2$

Answer 1

$f\left(x\right)=x^5+2x^3+3x+4$

$f^{\prime }\left(x\right)=5x^4+6x^2+3$

$\frac{d}{dx}\left(f^{\prime }\left(x\right)\right)=20x^3+12x$

$f^{\prime \prime }\left(x\right)=20x^3+12x$ at $x=-2$ is

$f^{\prime \prime }(-2)=20\times {(-2)}^3+12\times -2$$=-160-24=-184$

$28\frac{d}{dx}\left(f^{\prime }(-2)\right)=28\times -184=-5152$