# Assume f and g are differentiable on their domains with h

Assume f and g are differentiable on their domains with $h\left(x\right)=\left(fog\right)\left(r\right)$. Suppose the equation of the line tangent to the graph of g at the point $\left(4,7\right)$ is$y=3x–5$ and the equation of the line tangent to the graph of f at $\left(7,9\right)$ is $y=-2x+23$.
(a) Calculate $h\left(4\right)$ and ${h}^{\prime }\left(4\right)$.
(b) Determine an equation of the line tangent to the graph of h at the point on the graph where $x=4$.
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Step 1:Define the functionsNSk(a)
Since $h\left(4\right)=f\left(g\left(4\right)\right)$ and ${h}^{\prime }\left(4\right)={f}^{\prime }\left(g\left(4\right)\right).{g}^{\prime }\left(4\right)$ , therefore, to determine the value of $h\left(4\right)$ and ${h}^{\prime }\left(4\right)$ : we need to know the values of $g\left(4\right),f\left(g\left(4\right)\right),{g}^{\prime }\left(4\right)$ and ${f}^{\prime }\left(g\left(4\right)\right)$ .
Now, $\left(4,7\right)$ point lies on the graph of g, therefore, $g\left(4\right)=7$.
Similarly, $\left(7,9\right)$ lies on graph f , therefore, $f\left(7\right)=9$.
Now, $g\left(4\right)=7$ , therefore, $f\left(g\left(4\right)\right)=9$.
We know that if a line is tangent to a graph at a point $\left(x,y\right)$, then differentiation of the function will have the same value to the slope of the line.
Now, slope of the line : $y=3x-5$ is 3.
As the the line $y=3x-5$ is tangent to the graph g on $\left(4,7\right)$, therefore, ${g}^{\prime }\left(4\right)=3$.
Similarly,
we have, slope of the line $y=-2x+23$ is -2 .
Therefore, ${f}^{\prime }\left(7\right)=-2$.
Therefore, ${f}^{\prime }\left(g\left(4\right)\right)=-2$.
So,
$h\left(4\right)=f\left(g\left(4\right)\right)=9$.
And,
${h}^{\prime }\left(4\right)={f}^{\prime }\left(g\left(4\right)\right).{g}^{\prime }\left(4\right)=-2.3=-6$.

Neil Dismukes
Step 2: Second Part
(b)
Using the formula of tangent , we have :
$y-h\left(4\right)={h}^{\prime }\left(4\right).\left(x-4\right)$ $\left[atx=4\right]$
$⇒y-9=\left(-6\right).\left(x-4\right)$
$⇒y=-6x+33$.
Therefore, equation of the tangent line is :
$y=-6x+33$.