find-the-derivatives-of-the-given-functions-ytan-1leftfrac1-t1tright-3

Question

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

Tyler Moulton · Accepted Answer

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&#x27;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&#x27;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&#x27;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.

Theresa Nguyen · Answer

or the derivative of why is equal to tangent t over one plus seeking to t The first thing you want to do is to apply be quotient. Rule F over G crime is equal to F Prime times G minus G prime times f All of this is over g squared The apple in application of this rule will look like the over d t of tangent of T multiplied by one plus seeking T minus D over DT of one plus seeking to t multiplied by tangent t All of this is over one plus seek and t squared. Now that we have this, let&#x27;s look for the derivative of tension tea and one plus seeking t so d over DT of Tangent of T is equal to seek and squared t next D over DT of one plus seeking T is equal to seek into t Tangent City. So know that we have these derivatives taken care of Let&#x27;s put it all back together we get second squared T multiplied by one plus seek and t minus seek it t well supplied by tangent t tangent ity. All of this is over one class second T squared case. I know that we have this. Let us simplify Sequent and Tangent with sign and co sign. This will look like one plus or actually won over. Toast and tea squared, multiplied by one plus one over coastline T minus one over coastline t multiplied by sine of t over coastline t squared All of this is over one plus one over co sign t squared And now that we have this, we will simplify even further. We&#x27;re first going to simplify this second and notice his third. So the first simplification one plus one over co sign of tea squared is going to be equal to co sign of t plus one squared over co sign squared t this this one So to who will be simplified as one over co sign of tea squared both supplied by one plus one over co sign of tea is equal to co sign T plus one over co sign cubed T last one. Number three we&#x27;re going to simplify one over co sign t multiplied by ah sine of t over ko 70 squared that is simplified into sine squared t over co sign cubed t putting all of these back together we will get co sign T plus one over co sign huge tea minus sine squared of tea over co sign cubed t. All of this is over. Co sign of t plus one squared over co sine squared t Okay, so our next step will be to combine our fractions. You notice Here we have co sign Q T and cosign Cube T This will look like co sign of t plus one minus sine squared and he over co sign cubed t. All of this is over. Co sign t plus one squared over co sign square to T. The next thing we want to do is to manage our fraction. A Overbey oversee over de is equal to a times D over b times c So this will look like co sign of t plus one minus sign squared T multiplied by co sine squared t All of this is over. Close sign cubed T multiplied by co sign of t plus one squared. The next thing we want to do is to apply our exponents. Rule X to the A over X to the B is equal to one over X. It&#x27;s the B minus a X to the A is co sine squared T ax to the B is co sign Q T. This will look like co sign squared T over coastline huge tea, which is equal to one over co sign of teeth, Putting it all back together we will get close Sign of tea plus one minus sine squared T over co signed T multiplied by co sign T Plus one squared. So the next step we&#x27;re going to use the identity of one minus sine squared X being equal to co sign Squared X The application of this rule well, look like co sign of tea plus co sign Squared T over one plus co sign of tea Squared Co sign T. The next thing we want to do is factor out. Co sign from the numerator. We will get co sign of tea multiplied by one plus co sign t. All of this is over one plus close sign of T squared multiplied by co sign T. The last step is to cancel Let&#x27;s cancel Coastline and co sign one plus co sign t one plus ko 70. So our final answer will be 1/1 plus co sign t

Ernest Castorena · Answer

from this time get one that repetitive of this function. Virginity is the caution role. We could take the derivative of the top and then multiply it by the bottom. And then we have the same track being at the top alone this time. But multiply it by that everybody over the bottom and, uh, all over one second square. Now we can simplify because we know that Let&#x27;s see sine squared co sign square is equal to one. We can divide everything by co sign and attention squared plus one as equal twos where so we can replace over yet and we&#x27;ll end up with tendon squared plus one times one minus Seacon square chicken tangent. So this is the numerator right now and let&#x27;s see how I can manipulate this. I could end up with attendant squared T can with one economic with a sickened tension squared on just a second T. Now this is going to cancel out with this. So we have a tension squared plus one. Yes, he can, but that engine squared is equal. Teoh seeking squared T minus one so we could plug that back and and we did a poll in a second squared Teoh minus one plus one plus seeking d. So that was all for a numerator. And our numerator is now all is congee gotten rid of? We can pull out the Sikh indeed. And we end up with one mind, uh, very seeking t plus one. That&#x27;s what we end up. He can t plus one because we pack it out over seek in plus one squared. But what happens here? The top cancels with one of the bottom. And so now we have made a lot of progress. What we have is one over co sign t over one over co sign, have tea and plus one so we can probably simplify it more if you&#x27;re wondering. So let&#x27;s get the bottom under a common denominator and we&#x27;ll get one plus co sign of t over co sign of tea. And then we can see that this cancels this and we end up with one over one plus co sanity

Amy Jiang · Answer

I think so. It&#x27;s find of primal T. So using our change roles, starting without letna deserve it of a bat that&#x27;s equal to one over inside term and in time determinative off our inside judge. It&#x27;s immersed, which is equal to 1/1 plus p squared. Okay, so we get one over engines. Inverse of kidney times one plus he squared for a solution.

Mike Gaerlan · Answer

in order to find the derivative of this vector value function here that it&#x27;s our lives, our prime of tea this is gonna be equal to. So we&#x27;ll take the derivative of the first component. So that&#x27;s just going to be will have negative t plus one to the negative, too. Uh, and then times that over the inside, which is just one. So just one. Now, the derivative of the inverse tension of tea, this is equal to 1/1 plus t squared. And then, lastly, the derivative of Ln is one over. And then, uh so that&#x27;s just gonna become one over t plus one. And then remember, chain rule, you think the derivative of this, which is as times one Okay, so that&#x27;s going to be our derivative after here.

Sriram Soundarrajan · Answer

Hello. Tamar could get right. That&#x27;s our problem about that day here function for 50 years ago by alert. Then it was off the So we have to find the director of this particular anybody. Long day Bannon was off Indo by not less the cloud.

Problem

Find the derivatives of the given functions. $$p=…

Problem 27 Easy Difficulty

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

Answer

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

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

Heather Z.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Basic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

Heather Z.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

Heather Z.

Kristen K.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus