Question

Differentiate y=e^[p](p+p \sqrt{p})

Answer 1

Answer:

$e^p(p+p\sqrt{p}+1+\frac{3}{2}\sqrt{p})$  

Step-by-step explanation:

We are given the following information in the question:

$y = e^p(p+p\sqrt{p})\\\\y = e^p(p + p^{\frac{3}{2}})$

We will use the product rule to differentiate the above expression.

The product rule says that:

$\displaystyle\frac{d(uv)}{dx} = u.\frac{dv}{dx} + v \frac{du}{dx}$

The differentiation is done in the following ways:

$\displaystyle\frac{dy}{dx} = \frac{d(e^p)}{dp}(p+p\sqrt{p}) + e^p\frac{d(p+p\sqrt{p})}{dp}\\\\= e^p(p+p\sqrt{p}) + e^p(1+\frac{3}{2}\sqrt{p})\\\\= e^p(p+p\sqrt{p}+1+\frac{3}{2}\sqrt{p})$

The above is the required differentiation of the given expression.

Answer 2

$y=e^p(p+p\sqrt p)=e^p(p+p^{3/2})$

$\dfrac{\mathrm dy}{\mathrm dp}=e^p(p+p^{3/2})+e^p\left(1+\dfrac32p^{1/2}\right)=e^p\left(1+\dfrac32p^{1/2}+p+p^{3/2}\right)$