How to Differentiate f(x) = x√(2x − 3)

To differentiate $f(x)=x\sqrt{2x-3}$, use the product rule because the expression is a product of $x$ and a square root. The square root also contains an inner function, $2x-3$, so differentiating it requires the chain rule.

The result is

$$f'(x)=\frac{3(x-1)}{\sqrt{2x-3}},$$

for $x>\frac32$. Here is how to obtain it and why the endpoint needs separate attention.

Apply the product rule

For a product $u(x)v(x)$, the product rule says

$$(uv)'=u'v+uv'.$$

Set $u(x)=x$ and $v(x)=\sqrt{2x-3}$. Then $u'(x)=1$. For the second factor, write it as $(2x-3)^{1/2}$ and apply the chain rule:

$$\begin{aligned} v'(x)&=\frac12(2x-3)^{-1/2}\cdot 2 \\ &=\frac{1}{\sqrt{2x-3}}. \end{aligned}$$

The factor of $2$ comes from differentiating the inside, $2x-3$. Substituting both derivatives into the product rule gives

$$\begin{aligned} f'(x)&=\sqrt{2x-3}+\frac{x}{\sqrt{2x-3}} \\ &=\frac{2x-3}{\sqrt{2x-3}}+\frac{x}{\sqrt{2x-3}} \\ &=\frac{3x-3}{\sqrt{2x-3}} \\ &=\frac{3(x-1)}{\sqrt{2x-3}}. \end{aligned}$$

The second line rewrites the first term with the same denominator as the second. This is the simplified derivative.

Check the domain

For the original function to be real-valued, the radicand must be nonnegative:

$$2x-3\geq 0,$$

so $f$ is defined for $x\geq\frac32$. But the derivative formula has $\sqrt{2x-3}$ in its denominator. It is therefore defined only when $2x-3>0$, or $x>\frac32$.

At $x=\frac32$, the function itself is defined and equals zero, but it has no finite right derivative. For small $h>0$, its difference quotient at that endpoint is

$$\frac{f(\frac32+h)-f(\frac32)}{h} =\frac{(\frac32+h)\sqrt{2h}}{h},$$

which grows without bound as $h$ approaches zero. Thus the endpoint is not included in the derivative’s domain.

Worked check at x = 2

The function value at $x=2$ is $f(2)=2\sqrt{1}=2$. The derivative formula gives

$$f'(2)=\frac{3(2-1)}{\sqrt{4-3}}=3.$$

Using the unsimplified product-rule expression gives the same result: $\sqrt{1}+2/\sqrt{1}=1+2=3$. So the tangent line at the point $(2,2)$ has slope $3$ and equation $y-2=3(x-2)$, or $y=3x-4$.

Common mistakes

• Using only the product rule or only the chain rule. The product rule handles the two factors; the chain rule handles the expression inside the square root. Both are needed.
• Forgetting the inner derivative. Differentiating $(2x-3)^{1/2}$ introduces a factor of $2$. It cancels with the exponent’s factor of $\frac12$ here, leaving $1/\sqrt{2x-3}$.
• Dropping the product-rule term. The derivative is not just $x/\sqrt{2x-3}$; the derivative of the first factor contributes $\sqrt{2x-3}$ as well.
• Including the endpoint in the derivative’s domain. The function is defined at $x=\frac32$, but the derivative expression is not, and the endpoint difference quotient does not approach a finite value.

For $x>\frac32$, the derivative can be written either as $\sqrt{2x-3}+x/\sqrt{2x-3}$ or in simplified form as $3(x-1)/\sqrt{2x-3}$.