`l_(1)=(d)/(dx)(e^(tanx))`
`l_(2)=lim_(h to 0)(e^(sin(x+h)-e^(sinx)))/(h)`
`l_(3)=inte^(sinx)cosxdx`, then which one of the following is correct ?

Options

A. `l_(1) ne l_(2)`
B. `(d)/(dx)(l_(3))=l_(2)`
C. `intl_(3)dx=l_(2)`
D. `l_(2)=l_(3)`

Answer

To solve the problem, we need to analyze the three expressions \( l_1 \), \( l_2 \), and \( l_3 \) given in the question.

### Step 1: Evaluate \( l_1 \)

Given:
\[ l_1 = \frac{d}{dx}(e^{\tan x}) \]

Using the chain rule for differentiation:
\[ \frac{d}{dx}(e^{u}) = e^{u} \cdot \frac{du}{dx} \]
where \( u = \tan x \).

Now, we find \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \sec^2 x \]

Thus,
\[ l_1 = e^{\tan x} \cdot \sec^2 x \]

### Step 2: Evaluate \( l_2 \)

Given:
\[ l_2 = \lim_{h \to 0} \frac{e^{\sin(x+h)} - e^{\sin x}}{h} \]

This limit represents the derivative of \( e^{\sin x} \) using the definition of the derivative:
\[ l_2 = \frac{d}{dx}(e^{\sin x}) \]

Using the chain rule again:
\[ \frac{d}{dx}(e^{\sin x}) = e^{\sin x} \cdot \cos x \]

Thus,
\[ l_2 = e^{\sin x} \cdot \cos x \]

### Step 3: Evaluate \( l_3 \)

Given:
\[ l_3 = \int e^{\sin x} \cos x \, dx \]

The integral of \( e^{\sin x} \cos x \) can be solved using substitution. Let:
\[ u = \sin x \]
Then,
\[ du = \cos x \, dx \]

Substituting into the integral gives:
\[ l_3 = \int e^{u} \, du = e^{u} + C = e^{\sin x} + C \]

### Conclusion

Now we summarize the results:
- \( l_1 = e^{\tan x} \cdot \sec^2 x \)
- \( l_2 = e^{\sin x} \cdot \cos x \)
- \( l_3 = e^{\sin x} + C \)

From the analysis, we can see that:
- The derivative of \( l_3 \) gives us \( l_2 \), which means \( l_2 \) is indeed the derivative of \( l_3 \).

Thus, the correct statement is that \( l_2 \) is the derivative of \( l_3 \).

### Final Answer
The correct option is: **Option 2**.