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Given y = f(x) = (1)/(sqrt(x+1)).

Mathematics

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Answer

−12(x+1)3/2-\frac{1}{2(x+1)^{3/2}}

Step-by-step solution

6 steps

  1. Write down the definition of the derivative from the first principle.

    The derivative of a function f(x)f(x) is given by:

    dydx=lim⁡h→0f(x+h)−f(x)h\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

  2. Identify f(x)f(x) and f(x+h)f(x+h).

    Given y=f(x)=1x+1y = f(x) = \frac{1}{\sqrt{x+1}}.
    Then f(x+h)=1(x+h)+1=1x+h+1f(x+h) = \frac{1}{\sqrt{(x+h)+1}} = \frac{1}{\sqrt{x+h+1}}.

  3. Substitute f(x)f(x) and f(x+h)f(x+h) into the limit definition.

    dydx=lim⁡h→01x+h+1−1x+1h\frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h+1}} - \frac{1}{\sqrt{x+1}}}{h}

  4. Simplify the numerator by finding a common denominator.

    dydx=lim⁡h→0x+1−x+h+1x+h+1x+1h\frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{\sqrt{x+1} - \sqrt{x+h+1}}{\sqrt{x+h+1}\sqrt{x+1}}}{h} dydx=lim⁡h→0x+1−x+h+1hx+h+1x+1\frac{dy}{dx} = \lim_{h \to 0} \frac{\sqrt{x+1} - \sqrt{x+h+1}}{h\sqrt{x+h+1}\sqrt{x+1}}

  5. Multiply the numerator and denominator by the conjugate of the numerator.

    The conjugate of x+1−x+h+1\sqrt{x+1} - \sqrt{x+h+1} is x+1+x+h+1\sqrt{x+1} + \sqrt{x+h+1}.

    dydx=lim⁡h→0x+1−x+h+1hx+h+1x+1×x+1+x+h+1x+1+x+h+1\frac{dy}{dx} = \lim_{h \to 0} \frac{\sqrt{x+1} - \sqrt{x+h+1}}{h\sqrt{x+h+1}\sqrt{x+1}} \times \frac{\sqrt{x+1} + \sqrt{x+h+1}}{\sqrt{x+1} + \sqrt{x+h+1}} dydx=lim⁡h→0(x+1)2−(x+h+1)2hx+h+1x+1(x+1+x+h+1)\frac{dy}{dx} = \lim_{h \to 0} \frac{(\sqrt{x+1})^2 - (\sqrt{x+h+1})^2}{h\sqrt{x+h+1}\sqrt{x+1}(\sqrt{x+1} + \sqrt{x+h+1})} dydx=lim⁡h→0(x+1)−(x+h+1)hx+h+1x+1(x+1+x+h+1)\frac{dy}{dx} = \lim_{h \to 0} \frac{(x+1) - (x+h+1)}{h\sqrt{x+h+1}\sqrt{x+1}(\sqrt{x+1} + \sqrt{x+h+1})} dydx=lim⁡h→0x+1−x−h−1hx+h+1x+1(x+1+x+h+1)\frac{dy}{dx} = \lim_{h \to 0} \frac{x+1-x-h-1}{h\sqrt{x+h+1}\sqrt{x+1}(\sqrt{x+1} + \sqrt{x+h+1})} dydx=lim⁡h→0−hhx+h+1x+1(x+1+x+h+1)\frac{dy}{dx} = \lim_{h \to 0} \frac{-h}{h\sqrt{x+h+1}\sqrt{x+1}(\sqrt{x+1} + \sqrt{x+h+1})}

  6. Cancel out hh and evaluate the limit.

    dydx=lim⁡h→0−1x+h+1x+1(x+1+x+h+1)\frac{dy}{dx} = \lim_{h \to 0} \frac{-1}{\sqrt{x+h+1}\sqrt{x+1}(\sqrt{x+1} + \sqrt{x+h+1})}

    Now, substitute h=0h=0:

    dydx=−1x+0+1x+1(x+1+x+0+1)\frac{dy}{dx} = \frac{-1}{\sqrt{x+0+1}\sqrt{x+1}(\sqrt{x+1} + \sqrt{x+0+1})} dydx=−1x+1x+1(x+1+x+1)\frac{dy}{dx} = \frac{-1}{\sqrt{x+1}\sqrt{x+1}(\sqrt{x+1} + \sqrt{x+1})} dydx=−1(x+1)(2x+1)\frac{dy}{dx} = \frac{-1}{(x+1)(2\sqrt{x+1})} dydx=−12(x+1)1(x+1)1/2\frac{dy}{dx} = \frac{-1}{2(x+1)^{1}(x+1)^{1/2}} dydx=−12(x+1)3/2\frac{dy}{dx} = \frac{-1}{2(x+1)^{3/2}}

    The derivative of y=1x+1y = \frac{1}{\sqrt{x+1}} from the first principle is −12(x+1)3/2\boxed{-\frac{1}{2(x+1)^{3/2}}}.

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