Question

Let $f,g:R\to R$ be two functions. If $4x-y-8=0$ and $12x+y+26=0$ are the tangents to the graph of the functions $f\left(x\right)$ and $g\left(x\right)$ at $x=1$ and $x=-3$ respectively, and $h\left(x\right)=(g(x+f(x\left)\right){)}^2,$ then $-\frac{h^{\prime }\left(1\right)}{100}$ is equal to

Answer 1

$h^{\prime }\left(x\right)=2g(x+f(x\left)\right)\cdot g^{\prime }(x+f(x\left)\right)\cdot (1+f^{\prime }(x\left)\right)$
Now, to find value of $f\left(x\right)$ at $x=1$
We know the tangent at $x=1$ on $f\left(x\right)$
$\Rightarrow 4\left(1\right)-y-8=0\Rightarrow y=-4=f\left(1\right)$
Slope of tangent, $f^{\prime }\left(1\right)=m_t=4$
$\Rightarrow h^{\prime }\left(1\right)=2g(1-4)\cdot g^{\prime }(1-4)\cdot (1+4)\Rightarrow h^{\prime }\left(1\right)=2g(-3)\cdot g^{\prime }(-3)\cdot 5$

For $g\left(x\right)$ at $x=-3,$
$12(-3)+y+26=0\Rightarrow y=10=g(-3)$
Slope of tangent, $m_t=g^{\prime }(-3)=-12$
$h^{\prime }\left(1\right)=2\left(10\right)(-12)\left(5\right)=-1200$
$\Rightarrow -\frac{h^{\prime }\left(1\right)}{100}=12$