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# Find the derivative of the function. Simplify where possible.$y = \sin^1 (2x + 1)$

## $y^{\prime}=\frac{1}{\sqrt{1-(2 x+1)^{2}}} \cdot \frac{d}{d x}(2 x+1)=\frac{1}{\sqrt{1-\left(4 x^{2}+4 x+1\right)}} \cdot 2=\frac{2}{\sqrt{-4 x^{2}-4 x}}=\frac{1}{\sqrt{-x^{2}-x}}$

Derivatives

Differentiation

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

Yeah. In this problem, we are learning how to use implicit differentiation. Specifically, when we have an inverse, trig and metric function, we are given the function. Why equals the inverse sign of two X plus one. Now, if you've looked at implicit differentiation before, you might be able to tell that one side of our equation is pretty easy to solve, we just have why on one side. So the first thing that we're going to do with implicit differentiation is take the derivative of each side of our equation. So we would have d y DX equals the derivative of the inverse sign of two X plus one. So do y DX were good on that. That is what we want to find our derivative now, to find the derivative of our inverse sine function, we have to use the chain rule. We have a composition of functions here, so we would get do I d. X equals 1/1 minus two x plus one squared times two times one plus zero. And this portion right here comes from the derivative of the inverse sign. And then we could simplify to get d Y d X equals 2/1 minus four X squared plus four x plus one That was just essentially foiling this, um, binomial here we could simplify again to get to over negative for X squared minus four x, and we're almost done. Just a little bit more simplification will get d Y d X equals 2/2 times negative X squared minus X And then we can cancel those those twos to get that are derivative. D Y D x is one over the square root of negative X squared minus X, and that is our derivative. I hope that this problem helped you understand a little bit more and about implicit differentiation specifically, how we can use implicit differentiation when we have an inverse trigger metric function involved in the function that were given.

University of Denver

#### Topics

Derivatives

Differentiation

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp