Question

If the three functions $f(x),g(x)$ and $h(x)$ are such that $h(x)=f(x),g(x)$ and $f^{\prime }(x)\cdot g^{\prime }(x)$ = c, where $c$ is a constant, then $\frac{f^{\prime \prime }(x)}{f(x)}+\frac{g^{\prime \prime }(x)}{g(x)}+\frac{2c}{f(x).g(x)}$ is equal to ____

• A. $\frac{h^{\prime \prime }(x)}{h(x)}$
• B. $\frac{h(x)}{h^{\prime }(x)}$
• C. $h^{\prime }(x)h^{\prime \prime }(x)$
• D. $\frac{h(x)}{h^{\prime \prime }(x)}$

Answer 1

Answer: (A)

Given, $h(x)=f(x)\cdot g(x)$ and $f^{\prime }(x)\cdot g^{\prime }(x)=c$
Now, $h^{\prime }(x)=f^{\prime }(x)\cdot g(x)+f(x)\cdot g^{\prime }(x)$
$h^{\prime \prime }(x)=f^{\prime \prime }(x)\cdot g(x)+f^{\prime }(x)\cdot g^{\prime }(x)$
$+f^{\prime }(x)\cdot g^{\prime }(x)+f(x)\cdot g^{\prime \prime }(x)$
$h^{\prime \prime }(x)=f^{\prime \prime }(x)\cdot g(x)+f(x)\cdot g^{\prime \prime }(x)$
$+2f^{\prime }(x)\cdot g^{\prime }(x)$
$h^{\prime \prime }(x)=f^{\prime \prime }(x)\cdot g(x)+f(x)\cdot g^{\prime \prime }(x)+2c$...(i)
Now, we find
$\frac{f^{\prime \prime }(x)}{f(x)}+\frac{g^{\prime \prime }(x)}{g(x)}+\frac{2c}{f(x)\cdot g(x)}$
$=\frac{f^{\prime \prime }(x)\cdot g(x)+g^{\prime \prime }(x)\cdot f(x)+2c}{f(x)\cdot g(x)}$
$=\frac{h^{\prime \prime }(x)}{h(x)}$ [from Eq. (i)]