Question

If $\frac{d}{dx}(f(x))=g(x)$ and $\frac{d}{dx}(g(x))=f\left(x^2\right)$ then $\frac{d^2}{d{x}^2}\left(f\left(x^3\right)\right)$ can be expressed in the form $k$ $x^{a}f\left(x^b\right)+pxg\left(x^c\right)$ where $a,b,c,k,p\in N$, then the value of $(k+p+a+b+c)$ equals

• A. 26
• B. 27
• C. 28
• D. 30

Answer 1

Answer: (C)

Given $f^{\prime }(x)=g(x)$
$g^{\prime }(x)=f\left(x^2\right)=f^{\prime \prime }(x)$
$\text{ let }y=f\left(x^3\right)$
$\frac{dy}{dx}=f^{\prime }\left(x^3\right)\cdot 3x^2$
$\frac{d^{2}y}{d{x}^2}=3\left[3x^2{x}^2\cdot f^{\prime \prime }\left(x^3\right)+2x\cdot f^{\prime }\left(x^3\right)\right]=9x^4{f}^{\prime \prime }\left(x^3\right)+6x{f}^{\prime }\left(x^3\right)=9x^4\cdot f\left(x^6\right)+6xg\left(x^3\right)$
$k{x}^{a}f\left(x^b\right)+pxg\left(x^c\right)$
$\therefore (k+p+a+b+c)=9+4+6+6+3=28$