Question

find dy for ( y = e^{sqrt{x}+1} ).
for ( y = e^{sqrt{x}+1} ), ( dy = ) dx.
(type an exact answer, using radicals as needed.)

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:

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