Question

I may you please help me with the diritives for AP Calculus

Answer 1

To find the derivative of a composite function we use the chain rule

$f(g(x))^{\prime}=f^{\prime}(g(x))g^{\prime}(x)$

Then, we first find the derivatives of f and g. The function f is of the form

$\begin{gathered} \frac{d}{dx}u^a=au^{a-1}\frac{du}{dx} \\ u=x^2-3 \\ a=\frac{1}{2} \\ \frac{du}{dx}=2x \end{gathered}$

Then

$f^{\prime}(x)=\frac{1}{2}(x^2-3)^{-1/2}(2x)=\frac{x}{\sqrt[]{x^2-3}}$

On the other hand, the derivative of g is very simple

$g^{\prime}(x)=4$

After doing this, we evaluate f' in g(x), and we have

$f^{\prime}(g(x))=\frac{4x-2}{\sqrt[]{(4x-2)^2-3}}$

Replacing f'(g(x)) and g'(x) we obtain

$f(g(x))^{\prime}=f^{\prime}(g(x))g^{\prime}(x)\text{ = }\frac{4x-2}{\sqrt[]{(4x-2)^2-3}}\cdot4$

Finally, we evaluate this expression in x=1

$(f\circ g)^{\prime}(1)=4\frac{4(1)-2}{\sqrt[]{(4(1)-2)^2-3}}=4\cdot\frac{2}{\sqrt[]{2^2-3}}=\frac{8}{1}=8$

Then, the answer is 8