Explanation Given: y = arctan ( arcsin x ) Concept Used: Chain rule of differentiation is: d d x f [ g ( x ) ] = d d g f ( g ) d g d x , where u is a differentiable function at x and f is a differentiable function at u ( x ) d d x ( sin − 1 x ) = 1 1 − x 2 , d d x ( tan − 1 x ) = 1 1 + x 2 Power rule of differentiation formula d d x x n = n x n − 1 Calculation: To calculate y ′ , where y = arctan ( arcsin x ) Differentiate both side of the equation with respect to x as y ′ = d d x [ arctan ( arcsin x ) ] Now use chain rule of differentiation to solve it further, as y ′ = d d x [ arctan ( arcsin x ) ] = 1 1 + ( arcsin x ) 2 d d x ( arcsin x ) = 1 1 +

15th Edition

ISBN: 9781305420878

Author: Stewart

Publisher: CENGAGE L

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Chapter 3, Problem 48E

To determine

To calculate y′ , where

y=arctan(arcsinx)

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Chapter 3, Problem 47E

Chapter 3, Problem 49E

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Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

Could you explain this using the formula I attached and polar coorindates

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Chapter 3 Solutions

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To calculate y ′ , where y = arctan ( arcsin x )

Solution Summary: The author explains how to calculate yprime, where u is a differentiable function at x, and f is

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To calculate y′ , where y=arctan(arcsinx)

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Chapter 3 Solutions

To calculate y′ , where

y=arctan(arcsinx)