Question

Find g′(4) given that f(4)=5, f′(4)=−1, and g(x)=(√x)*f(x).

Answer 1

Given that:

`g(x)=\sqrt[]{x}f(x)`

You need to find:

`g^(\prime)(x)`

In order to derivate the function, you need to apply the Product Rule

`(d)/(dx)(u\cdot v)=u\cdot v^(\prime)+v\cdot u^(\prime)`

Then, you get:

`g^(\prime)(x)=\sqrt[]{x}\cdot f^(\prime)(x)+f(x)(\sqrt[]{x})^(\prime)`

Since:

`\sqrt[]{x}=x^{(1)/(2)}`

You know that:

`(d)/(dx)(\sqrt[]{x})=(1)/(2)x^{(1)/(2)-1}=(1)/(2)x^{-(1)/(2)}=\frac{1}{2\sqrt[]{x}}`

Hence:

`\begin{gathered} g^(\prime)(x)=\sqrt[]{x}\cdot f^(\prime)(x)+f(x)(\frac{1}{2\sqrt[]{x}}) \\ \\ g^(\prime)(x)=\sqrt[]{x}\cdot f^(\prime)(x)+\frac{1}{2\sqrt[]{x}}f(x) \end{gathered}`

Knowing that you need to find:

`g^(\prime)(4)`

You can rewrite the function as follows:

`g^(\prime)(4)=\sqrt[]{4}\cdot f^(\prime)(4)+\frac{1}{2\sqrt[]{4}}f(4)`

Knowing that:

`\begin{gathered} f\mleft(4\mright)=5 \\ f^(\prime)\mleft(4\mright)=-1 \end{gathered}`

You can substitute values:

`g^(\prime)(4)=(\sqrt[]{4})(-1)+(\frac{1}{2\sqrt[]{4}})(5)`

Evaluating, you get:

`\begin{gathered} g^(\prime)(4)=(2)(-1)+((1)/(2\cdot2))(5) \\ \\ g^(\prime)(4)=-(3)/(4) \end{gathered}`

Hence, the answer is:

`g^(\prime)(4)=-(3)/(4)`