If u is a differentiable function of x, what is d/dx of e^{u(x)}?

If u is a differentiable function of x, what is d/dx of e^{u(x)}?

• A. e^{u(x)} • u'(x)
• B. e^{u(x)} / u'(x)
• C. u(x) e^{u(x)}
• D. e^{u'(x)}

Answer: (A)

The main idea here is the chain rule for the exponential function: when you have e raised to an inner function u(x), you multiply by the derivative of that inner function. Let y = e^{u(x)}. Taking natural logs gives ln y = u(x). Differentiating implicitly: (1/y) dy/dx = u'(x). Multiply both sides by y to get dy/dx = y · u'(x) = e^{u(x)} · u'(x). So the derivative is e^{u(x)} times u'(x). This uses that u is differentiable, so u'(x) exists, and the chain rule tells us to multiply the derivative of the outer function (which is 1 on the exponent, since d/dt e^t = e^t) by the derivative of the inner function. The other forms would mix up where the derivative goes or treat the exponent incorrectly (for example, dividing by u'(x), or replacing the exponent with u'(x), which doesn’t reflect the chain rule).

The main idea here is the chain rule for the exponential function: when you have e raised to an inner function u(x), you multiply by the derivative of that inner function.

Let y = e^{u(x)}. Taking natural logs gives ln y = u(x). Differentiating implicitly: (1/y) dy/dx = u'(x). Multiply both sides by y to get dy/dx = y · u'(x) = e^{u(x)} · u'(x).

So the derivative is e^{u(x)} times u'(x). This uses that u is differentiable, so u'(x) exists, and the chain rule tells us to multiply the derivative of the outer function (which is 1 on the exponent, since d/dt e^t = e^t) by the derivative of the inner function.

The other forms would mix up where the derivative goes or treat the exponent incorrectly (for example, dividing by u'(x), or replacing the exponent with u'(x), which doesn’t reflect the chain rule).