Find f'(x) given f(x) = \sqrt {1 + \sqrt{1 + x^2}} a. \frac x 2 (1 + (1 + x^2)^{1/2})^{-1/2} b....

Question:

Find {eq}f'(x) {/eq} given {eq}f(x) = \sqrt {1 + \sqrt{1 + x^2}} {/eq}

a. {eq}\frac x 2 (1 + (1 + x^2)^{1/2})^{-1/2} {/eq}

b. {eq}\frac 12 ( 1 + (1 + x^2)^{1/2})^{-1/2} (1 + x^2)^{-1/2} {/eq}

c. {eq}\frac x2 ( 1 + (1 + x^2)^{1/2})^{-1/2} (1 + x^2)^{-1/2} {/eq}

d. {eq}x ( 1 + (1 + x^2)^{1/2})^{-1/2} (1 + x^2)^{-1/2} {/eq}

e. None of the above

Double Square Root:


Suppose we have double square root function such as {eq}\sqrt{1+\sqrt{h(t)}} {/eq} and we need the first derivative of this function with respect to the variable t, then we'll rewrite the square root function as fractional exponent function and after this process, we'll compute the derivative of the whole function with respect to the function {eq}1+\sqrt{h(t)} {/eq} and then compute the derivative of this function {eq}1+\sqrt{h(t)} {/eq} with respect to the variable t (it is obtained frm the chain rule of derivatives).

Here, we have:

{eq}\displaystyle \frac{\mathrm{d} (1+\sqrt{h(t)})^{\frac{1}{2}}}{\mathrm{d} t}=\frac{\mathrm{d} (1+\sqrt{h(t)})^{\frac{1}{2}}}{\mathrm{d} (1+\sqrt{h(t)})}\cdot \frac{\mathrm{d} (1+\sqrt{h(t)})}{\mathrm{d} t} {/eq}

Answer and Explanation: 1

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Given data:

The double square root function is:

{eq}f(x) = \sqrt {1 + \sqrt{1 + x^2}} {/eq}

The fractional exonent function from the above...

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Derivatives: The Formal Definition

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Chapter 7 / Lesson 5
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The derivative in calculus is the rate of change of a function. In this lesson, explore this definition in greater depth and learn how to write derivatives.


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