Question

3-34:Differentiate the function.

34. \(y = {e^{x + 1}} + 1\)

Short answer

The derivative of given function is \(\frac{{dy}}{{dx}} = {e^{x + 1}}\).

Step-by-step solution

Different rules of differentiation

The given function is \(y = {e^{x + 1}} + 1\).

The sum rule is:\(\frac{d}{{dx}}\left( {g\left( x \right) + f\left( x \right)} \right) = \frac{d}{{dx}}\left( {g\left( x \right)} \right) + \frac{d}{{dx}}\left( {f\left( x \right)} \right)\)

And, the power rule is: \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\)

Evaluating Derivative using all these rules

Differentiating with respect to\(x\)as:

\(\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {{e^{x + 1}} + 1} \right)\\ &= \frac{d}{{dx}}\left( {{e^{x + 1}}} \right) + \frac{d}{{dx}}\left( 1 \right)\\ &= \frac{d}{{dx}}\left( {e \cdot {e^x}} \right) + 0\\ &= e \cdot \frac{d}{{dx}}\left( {{e^x}} \right)\\ &= e \cdot {e^x}\\ &= {e^{x + 1}}\end{align}\)

Hence, the derivative of given function is \(\frac{{dy}}{{dx}} = {e^{x + 1}}\).