h′(x)=2g(x+f(x))⋅g′(x+f(x))⋅(1+f′(x))
Now, to find value of f(x) at x=1
We know the tangent at x=1 on f(x)
⇒4(1)−y−8=0⇒y=−4=f(1)
Slope of tangent, f′(1)=mt=4
⇒h′(1)=2g(1−4)⋅g′(1−4)⋅(1+4)⇒h′(1)=2g(−3)⋅g′(−3)⋅5
For g(x) at x=−3,
12(−3)+y+26=0⇒y=10=g(−3)
Slope of tangent, mt=g′(−3)=−12
h′(1)=2(10)(−12)(5)=−1200
⇒−h′(1)100=12