Let y = 9√x + 5x - 1. Find dfracdydx.

Rewrite y = 9x^1/2 + 5x - 1. Differentiate term by term using the sum/difference rule: dy/dx = d/dx[9x^1/2] + d/dx[5x] - d/dx[1]. For the first term, apply the constant multiple rule and power rule: d/dx[9x^1/2] = 9 · 1/2x^-1/2 = 9/2x^-1/2. For the second term, d/dx[5x] = 5, and the derivative of the constant -1 is 0. Combining these gives dy/dx = frac92√x + 5.

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Derivative of a sum with a square root and linear term

Let $y = 9\sqrt{x} + 5x - 1$ . Find $\dfrac{dy}{dx}$ .

Hint

Rewrite $\sqrt{x}$ as $x^{1/2}$ , then differentiate each term separately using the power rule, applying the constant multiple rule to the coefficients and the sum/difference rule across terms. Remember the derivative of a constant is 0.

Step-by-step solution

Rewrite $y = 9x^{1/2} + 5x - 1$ . Differentiate term by term using the sum/difference rule: $\frac{dy}{dx} = \frac{d}{dx}[9x^{1/2}] + \frac{d}{dx}[5x] - \frac{d}{dx}[1]$ .

For the first term, apply the constant multiple rule and power rule: $\frac{d}{dx}[9x^{1/2}] = 9 \cdot \frac{1}{2}x^{-1/2} = \frac{9}{2}x^{-1/2}$ . For the second term, $\frac{d}{dx}[5x] = 5$ , and the derivative of the constant $-1$ is $0$ . Combining these gives $\frac{dy}{dx} = \frac{9}{2\sqrt{x}} + 5$ .

Derivative of a sum with a square root and linear term

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