To differentiate y = (√(x)−1)³ with respect to x, we will use the chain rule. Step 1: Let u = √(x)−1. Then y = u³. Step 2: Differentiate u with respect to x. We can write √(x) as x^(1/2). du/dx = d/dx(x^(1/2)−1) = 1/2x^((1/2)−1) − 0 = 1/2x^(−1/2) = (1)/(2√(x)) Step 3: Differentiate y with respect to u. dy/du = d/du(u³) = 3u²; Step 4: Apply the chain rule, which states dy/dx = dy/du · du/dx. dy/dx = (3u²) · ((1)/(2√(x))); Step 5: Substitute u = √(x)−1 back into the expression. dy/dx = 3(√(x)−1)² · (1)/(2√(x)) Step 6: Simplify the expression. dy/dx = (3(√(x)−1)²)/(2√(x)) The derivative is: (3(√(x)−1)²)/(2√(x))

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Let u = sqrt(x)−1.

May 21, 2026Mathematics

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Answer

\frac{3(\sqrt{x}-1)^2}{2\sqrt{x}}

Step-by-step solution

6 steps

To differentiate $y = (\sqrt{x}-1)^3$ with respect to $x$ , we will use the chain rule.

Let $u = \sqrt{x}-1$ .
Then $y = u^3$ .
Differentiate $u$ with respect to $x$ .
We can write $\sqrt{x}$ as $x^{1/2}$ .
$\frac{du}{dx} = \frac{d}{dx}(x^{1/2}-1) = \frac{1}{2}x^{(1/2)-1} - 0 = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$
Differentiate $y$ with respect to $u$ .
$\frac{dy}{du} = \frac{d}{du}(u^3) = 3u^2$
Apply the chain rule, which states $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .
$\frac{dy}{dx} = (3u^2) \cdot \left(\frac{1}{2\sqrt{x}}\right)$
Substitute $u = \sqrt{x}-1$ back into the expression.
$\frac{dy}{dx} = 3(\sqrt{x}-1)^2 \cdot \frac{1}{2\sqrt{x}}$
Simplify the expression.
$\frac{dy}{dx} = \frac{3(\sqrt{x}-1)^2}{2\sqrt{x}}$
The derivative is:
$\frac{3(\sqrt{x}-1)^2}{2\sqrt{x}}$

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Let u = sqrt(x)−1.

Answer

Step-by-step solution

Let u=x−1u = \sqrt{x}-1u=x​−1.

Differentiate uuu with respect to xxx.

Differentiate yyy with respect to uuu.

Apply the chain rule, which states dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​.

Substitute u=x−1u = \sqrt{x}-1u=x​−1 back into the expression.

Simplify the expression.

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More Mathematics Questions

Let $u = \sqrt{x}-1$ .

Differentiate $u$ with respect to $x$ .

Differentiate $y$ with respect to $u$ .

Apply the chain rule, which states $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

Substitute $u = \sqrt{x}-1$ back into the expression.