AP Calculus AB and BC

Derivative of e^sqrt(x): Answer, Proof, Mistakes

Q: What is the derivative of e^sqrt(x)?

It is \frac{e^{\sqrt{x}}}{2\sqrt{x}} for x>0, the original function divided by 2\sqrt{x}.

Q: Why does the derivative blow up at x = 0?

Because 2\sqrt{x} tends to 0 while e^{\sqrt{x}} tends to 1. The function itself is fine at x=0, with value 1, but its tangent line there is vertical.

Q: Can I do this with logarithmic differentiation?

Yes. Put y=e^{\sqrt{x}}, so \ln y=\sqrt{x} and \frac{y'}{y}=\frac{1}{2\sqrt{x}}. Multiplying back by y returns \frac{e^{\sqrt{x}}}{2\sqrt{x}}.

The derivative of e^sqrt(x) is e^sqrt(x) divided by 2 sqrt x. The exponential reproduces itself under differentiation, so the entire calculation reduces to the chain factor, and the derivative of sqrt x is 1 over 2 sqrt x, which divides rather than multiplies.

\frac{d}{dx}\left[e^{\sqrt{x}}\right] = \frac{e^{\sqrt{x}}}{2\sqrt{x}}

Differentiating e^sqrt(x)

Nothing about the exponential changes when you differentiate it, so all of the work sits in the inner function. Take $u=\sqrt{x}$ , so that $u'=\frac{1}{2\sqrt{x}}$ , and apply $\frac{d}{dx}e^{u}=e^{u}u'$ .

\frac{d}{dx}e^{\sqrt{x}} = e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}

The exponent stays exactly as it was. A derivative such as $e^{\sqrt{x}-1}$ or $e^{\frac{1}{2\sqrt{x}}}$ would mean the exponential rule had been mixed up with the power rule.

A tame function with an unbounded derivative

At $x=0$ the function is perfectly well behaved: $e^{\sqrt{0}}=1$ , and $e^{\sqrt{x}}$ is continuous on $[0,\infty)$ with no kink or gap. The derivative tells a different story.

\lim_{x\to 0^{+}}\frac{e^{\sqrt{x}}}{2\sqrt{x}} = +\infty

The numerator tends to $1$ and the denominator tends to $0$ from above, so the slope runs off to infinity and the graph leaves the point $(0,1)$ with a vertical tangent. Continuity at a point promises nothing about differentiability there, and this is a clean example of the gap.

Far to the right the derivative is large again, for the opposite reason: $e^{\sqrt{x}}$ eventually outgrows $2\sqrt{x}$ by any margin. The smallest slope sits in between, at $x=1$ , where the derivative equals $\frac{e}{2}$ .

The mistakes students make

Two of these come from stopping too early, and one from reaching for the wrong rule.

Answering $e^{\sqrt{x}}$ on its own. The exponential does reproduce itself, but the chain rule still demands the factor $\frac{1}{2\sqrt{x}}$ .
Answering $\frac{e^{\sqrt{x}}}{\sqrt{x}}$ and losing the $2$ . The derivative of $x^{1/2}$ is $\frac{1}{2}x^{-1/2}$ , and that one half is part of the answer.
Answering $\sqrt{x}\,e^{\sqrt{x}-1}$ by pulling the exponent down. The power rule applies to $x$ raised to a constant, not to a constant raised to a function of $x$ .

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of e^sqrt(x)?

It is $\frac{e^{\sqrt{x}}}{2\sqrt{x}}$ for $x>0$ , the original function divided by $2\sqrt{x}$ .

Why does the derivative blow up at x = 0?

Because $2\sqrt{x}$ tends to $0$ while $e^{\sqrt{x}}$ tends to $1$ . The function itself is fine at $x=0$ , with value $1$ , but its tangent line there is vertical.

Can I do this with logarithmic differentiation?

Yes. Put $y=e^{\sqrt{x}}$ , so $\ln y=\sqrt{x}$ and $\frac{y'}{y}=\frac{1}{2\sqrt{x}}$ . Multiplying back by $y$ returns $\frac{e^{\sqrt{x}}}{2\sqrt{x}}$ .