Evaluate dy/dx if `y=e^(sinx^2)`

Options

A. `2xcosx^2e^(sinx^2)`
B. `2xsinx^2e^(cosx^2)`
C. `2x^2e^(cosx^2)`
D. `x^2cosx^2e^(sinx^2)`

Answer

To evaluate \(\frac{dy}{dx}\) for the function \(y = e^{\sin(x^2)}\), we will use the chain rule of differentiation. Here are the steps to solve the problem:

### Step 1: Identify the outer and inner functions
In the function \(y = e^{\sin(x^2)}\), we can identify:
- The outer function \(g(u) = e^u\) where \(u = \sin(x^2)\)
- The inner function \(f(x) = \sin(x^2)\)

### Step 2: Differentiate the outer function
The derivative of the outer function \(g(u) = e^u\) with respect to \(u\) is:
\[ \frac{dg}{du} = e^u \]

### Step 3: Differentiate the inner function
Now we need to differentiate the inner function \(f(x) = \sin(x^2)\). We will again use the chain rule here:
- The derivative of \(\sin(v)\) is \(\cos(v)\) where \(v = x^2\)
- The derivative of \(v = x^2\) is \(2x\)

Thus, using the chain rule:
\[ \frac{df}{dx} = \cos(x^2) \cdot \frac{dv}{dx} = \cos(x^2) \cdot 2x = 2x \cos(x^2) \]

### Step 4: Apply the chain rule
Now we can apply the chain rule to find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{dg}{du} \cdot \frac{df}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = e^{\sin(x^2)} \cdot (2x \cos(x^2)) \]

### Final Result
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = 2x \cos(x^2) e^{\sin(x^2)} \]

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