Assuming that this function is differentiable, solve {eq}y = \frac{1 + xf(x)}{\sqrt {x}}. {/eq}

## Question:

Assuming that this function is differentiable, solve {eq}y = \frac{1 + xf(x)}{\sqrt {x}}. {/eq}

## Derivative:

The rate of change for an independent variable known as derivative. The derivative decreases the degree of the function. Apply the power rule and quotient rule:

{eq}\begin{align*} \frac{\mathrm{d} (\frac{u}{v})}{\mathrm{d} x}&=\frac{(\frac{\mathrm{d} u}{\mathrm{d} x}).v-u.(\frac{\mathrm{d} v}{\mathrm{d} x})}{v^{2}}\\ \frac{\mathrm{d} x^{n}}{\mathrm{d} x}&=nx^{n-1}\\ \end{align*} {/eq}.

## Answer and Explanation: 1

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View this answerGiven a function {eq}y = \frac{1 + xf(x)}{\sqrt {x}} {/eq}.

{eq}\begin{align*} y &= \frac{1 + xy}{\sqrt {x}}\\ y\sqrt {x}&=1+xy\\ y(\sqrt...

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Chapter 7 / Lesson 5The 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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