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} (d)/(dx)u^a=au^(a-1)(du)/(dx) \\ u=x^2-3 \\ a=(1)/(2) \\ (du)/(dx)=2x \end{gathered}`

Then

`f^(\prime)(x)=(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}}=(8)/(1)=8`

Then, the answer is 8