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Find the derivatives of the given functions.$$y=\tan ^{-1}\left(\frac{1-t}{1+t}\right)$$

$\frac{d y}{d t}=-\frac{1}{1+t^{2}}$

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 3

Derivatives of the Inverse Trigonometric Functions

Derivatives

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University of Michigan - Ann Arbor

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Yeah, we want to find Dy DT the derivative. The wives perspective team for y equals arc tangent of one dynasty over one plus two. To solve this problem, we're going to rely on derivative shortcuts. We picked up through a single variable calculus in particular are recently learned inverse trig and metric derivatives. So I have listed here the three major inverse trig derivatives. We need to use the D S r t n X. That was one of the one plus X square. As I've noted, we also need to use the chain rule as well as for the quotient rule for one I S. C over one plus two. Why has already written it's most easily differential form. Thus we can proceed to solve. So do I. D T is equal to negative 2/1 plus two square. This is from the derivative arc tangent. Yeah, divided by one plus one minus T over one plus Z squared. So in the numerator we have from the chain rule derivative 1/1 plus T. In the denominator one plus X squared. For the arguments are tangent. Simplifying give final solution. Dy DT is equal to negative 1/1 plus T squared.

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