find dy for ( y = e^{sqrt{x}+1} ).\nfor ( y = e^{sqrt{x}+1} ), ( dy = ) dx.\n(type an exact answer, using…

Explanation:

Step1: Let ( u = \sqrt{x}+1 )

Then ( y = e^{u} ).

Step2: Find ( \frac{du}{dx} )

( \frac{du}{dx}=\frac{1}{2\sqrt{x}} ) (using the power rule ( \frac{d}{dx}(x^{n})=nx^{n - 1} ), here ( n=\frac{1}{2} )).

Step3: Find ( \frac{dy}{du} )

Since ( y = e^{u} ), ( \frac{dy}{du}=e^{u} ).

Step4: Use the chain rule ( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} )

Substitute ( u=\sqrt{x}+1 ), ( \frac{dy}{du}=e^{u} ) and ( \frac{du}{dx}=\frac{1}{2\sqrt{x}} ) into the chain - rule formula. ( \frac{dy}{dx}=e^{\sqrt{x}+1}\cdot\frac{1}{2\sqrt{x}} ).

Step5: Find ( dy )

Since ( dy=\frac{dy}{dx}dx ), then ( dy=\frac{e^{\sqrt{x}+1}}{2\sqrt{x}}dx ).

Answer:

( \frac{e^{\sqrt{x}+1}}{2\sqrt{x}} )