find-the-derivatives-of-the-given-functions-v04-u-tan-1-2-u-3

Question

Find the derivatives of the given functions.
$$v=0.4 u 	an ^{-1} 2 u$$

Tyler Moulton · Accepted Answer

All right. We want to find the derivative dy dx or the function Y equals 0.4 X. Arc tangent of two. X. We&#x27;re going to use derivatives shortcuts that we&#x27;ve learned through a single variable calculus to solve this problem. In particular, we&#x27;re going to rely on the inverse trig functions or rather the inverse trig derivatives we recently learned. So the universe irregularities are given here. We need to rely on this problem on the D. S. R. 10 action was one of 11 sex where we also need to use the both the chain rule and the product will to solve. So why is already written in its most easily differentiable form? We start this derivative by taking the product rule. First point for X and then the derivative arc tangent to X. And then we take this up. So we have do I. D X equals 0.4. Our tandem two X 13.4 Access 0.4 plus 0.4 X times 2/1 plus four X square. This is the derivative of arc tangent to X, thus simplifying the expression we have. DY dx is equal 2.4 arc tangent of two X plus point X over one plus four X squared.

Jeffrey Utley · Answer

this question asked us to find the derivative of a given function. To start off, we must will get the function itself. F of y is equal to tangent in verse of two Y squared minus four. To evaluate this derivative, we must use the chain rule. Since we have a function inside the parentheses to I squared minus four. But we start with the derivative of tangent in verse. Tangent, inverse of something has derivative one over one. Plus whatever is in the parentheses squared. What we have is two y squared, minus four. So this is what&#x27;s gonna be in our pregnancies. Sorry about that one. There we go. Now we&#x27;re not done yet because we must multiply by the derivative of what is inside the parentheses. So the derivative of two y squared moving the two down By using the power rule, we arrive at four. Why? Since we multiply the twos and then we subtract one from the exponents who he&#x27;s just one taking the derivative of negative four will be zero. Since it is a constant so negative. I&#x27;m sorry. So four, Why will be our do it? It is so to simplify this down all we really need to do is put the four y on top. If we wanted to, we could boil out this by no meal here. However, this would not be simples form because it would be much more messy in the denominator. So our derivative of tangent inverse of two y squared minus four is equal to four y over one plus the quantity of two wife.

Amy Jiang · Answer

I think so. Fine. Why? We&#x27;ll start by taking our interpretive of tangent universe. The tension in verse is 1/1 plus R inside squared and then times the determinative of our inside terms. That&#x27;s four y. So we get or why over one waas to why squared among its four to park too. Okay, so this is our solution.

Shumayal N · Answer

So we&#x27;re gonna use changeable for this question and because our equation already has you and I&#x27;m gonna to note the inner term using the V instead of you. So our inner term is one plus to 10 U and our outer function f V or why is the to the power 4.5 Now, in order to apply the Changle first, we take the derivative of the outer function using the power. So that is gonna give us four. Bring five p to the power 3.5 and then we multiply that with the jirga tive of the inter function. So derivative of one is zero and the derivative of 2 10 year was going to be too second. She didn&#x27;t square, do you? And we substitute v ah into the equation to get the final answer. So 4.5 into two will give us nine and V is one plus 2 10 you to the power 3.5 multiplied with second square. You

Todd Vawdrey · Answer

Okay, So this problem, we&#x27;re finding the derivative of the tan squared X. So we&#x27;re gonna have to go ahead and apply the chain rule right here. Structurally, we have the tangents of ex inside of a squared function. So we&#x27;re gonna have to first do the squared. I&#x27;m using the power rule. So the two drops down and then we keep what the input waas. And then we need to multiply it by the inside of the function, which is that tan X and derivative of the tangent Is the C camp squared so altogether the derivative is to 10 x c can&#x27;t squared X.

Amy Jiang · Answer

Okay, so before we, um, take the derivative of the following will start for rewriting this as negative exponents, sending attention in verse of X squared. What&#x27;s for your fire of negative blood and I&#x27;ll taking our evidence. We&#x27;ll start with our power rule. That&#x27;s negative. One vengeance inverse of X squared plus four but a par of negative one on minus one said It&#x27;s negative, too. And then times derivative of tangent interest. That&#x27;s one over one plus R Inside squared and in times conservative of X squared plus four. That&#x27;s okay, so we get negative. Two x over. We can bring this sound, too or did not mean, but it&#x27;s vengeance in verse of squares was four to the power of two times one plus X squared plus four Part two.

Problem

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

Problem 17 Easy Difficulty

Find the derivatives of the given functions.
$$v=0.4 u \tan ^{-1} 2 u$$

Answer

$\frac{d v}{d u}=\frac{0.8 u}{1+4 u^{2}}+0.4 \tan ^{-1} 2 u$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Caleb E.

Kristen K.

Michael J.

Joseph L.

Precalculus Review - Intro

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$\frac{d v}{d u}=\frac{0.8 u}{1+4 u^{2}}+0.4 \tan ^{-1} 2 u$

Basic Technical Mathematics with Calculus

Caleb E.

Kristen K.

Michael J.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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

Caleb E.

Kristen K.

Michael J.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus