Find the derivative of y with respect to the appropriate variable. y = 4 \ln{x} + \sqrt{1 - x^2}...

Question:

Find the derivative of y with respect to the appropriate variable.

{eq}\displaystyle y = 4 \ln{x} + \sqrt{1 - x^2} \sec h^{-1}{x} {/eq}

Differentiation

This question is from the differentiation and we have to find out the derivative of the given function. We will use these formulae to solve this,

{eq}\Rightarrow \ \frac{d}{dx}(u\cdotp{v})=u\frac{d}{dx}v+u\frac{d}{dx}v\\ \Rightarrow \ \frac{d}{dx}(sech^{-1}(x))=\frac{1}{x\sqrt{x^{2}-1}}\\ {/eq}

Answer and Explanation:

{eq}\Rightarrow \ y=4\ln{x}+\sqrt{1-x^{2}}sech^{-1}(x)\\ \Rightarrow \ y'=\frac{d}{dx}(\ln{x}+\sqrt{1-x^{2}}sech^{-1}(x))\\ \Rightarrow \ y'=\frac{4}{x}+\sqrt{1-x^{2}}\frac{d}{dx}sech^{-1}(x))+sech^{-1}(x))\frac{d}{dx}\sqrt{1-x^{2}}\\ \Rightarrow \ y'=\frac{4}{x}+\frac{\sqrt{1-x^{2}}}{x\sqrt{x^{2}-1}}-\frac{xsech^{-1}(x)}{\sqrt{1-x^{2}}}\\ {/eq}


Learn more about this topic:

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Solving Partial Derivative Equations

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 1
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