Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 49

Answer

The equation for the tangent line is: $$y=\frac{7}{2}x-\frac{3}{2}.$$

Work Step by Step

The equation for the tangent line at $x_0$ is: $$y-f(x_0)=f'(x_0)(x-x_0)$$ Here is $x_0=1$ and $f(x)=x^2\sqrt{5-x^2}.$ $$f'(x)=(x^2\sqrt{5-x^2})'=(x^2)'\sqrt{5-x^2}+x^2(\sqrt{5-x^2})'= 2x\sqrt{5-x^2}+x^2\frac{1}{2\sqrt{5-x^2}}(5-x^2)'= 2x\sqrt{5-x^2}+x^2\frac{1}{2\sqrt{5-x^2}}\cdot(-2x)= 2x\sqrt{5-x^2}-\frac{x^3}{\sqrt{5-x^2}}=\frac{2x(5-x^2)-x^3}{\sqrt{5-x^2}}=\frac{10x-3x^3}{\sqrt{5-x^2}}$$ For $x=1$ we have: $$f(1)=1^2\sqrt{5-1^2}=1\cdot\sqrt{4}=2$$ $$f'(1)=\frac{10\cdot1-3\cdot1^3}{\sqrt{5-1^2}}=\frac{7}{2}$$ So, the equation for the tangent line is: $$y-2=\frac{7}{2}(x-1)\Rightarrow y=\frac{7}{2}x-\frac{7}{2}+2\Rightarrow y=\frac{7}{2}x-\frac{3}{2}$$
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