Question

Let $f,g$ and $h$ are differentiable functions. If $f\left(0\right)=1;g\left(0\right)=2;h\left(0\right)=3$ and the derivative of their pair wise products at $x=0$ are $\left(fg{)}^{\prime }\right(0)=6;(gh{)}^{\prime }\left(0\right)=4$ and $\left(hf{)}^{\prime }\right(0)=5$ then compute the value of $\left(fgh{)}^{\prime }\right(0).$

• A. $12$
• B. $15$
• C. $16$
• D. None of these

Answer 1

Answer: (C)

$16$

$y=fgh$

$\Rightarrow \frac{dy}{dx}=f^{\prime }gh+f{g}^{\prime }h+fg{h}^{\prime }=\frac{1}{2}(2f^{\prime }gh+2f{g}^{\prime }h+2fg{h}^{\prime })$

$=\frac{1}{2}(h(f^{\prime }g+g^{\prime }f)+g(f^{\prime }h+f{h}^{\prime })+f(g^{\prime }h+g{h}^{\prime }))$

$=\frac{1}{2}(h.{\left(fg\right)}^{\prime }+g.{\left(fh\right)}^{\prime }+f.{\left(gh\right)}^{\prime })$

$\therefore {\left(fgh\right)}^{\prime }\left(0\right)=\frac{1}{2}(h\left(0\right){\left(fg\right)}^{\prime }\left(0\right)+g\left(0\right){\left(fh\right)}^{\prime }\left(0\right)+f\left(0\right){\left(gh\right)}^{\prime }\left(0\right))$

$=\frac{1}{2}(3\times 6+2\times 5+1\times 4)=\frac{1}{2}(18+10+4)=\frac{32}{2}=16$