128. Derive the formula dy/dx = 1/(1+x²) for the derivative of...

Question

128. Derive the formula dy/dx = 1/(1+x²) for the derivative of y = arctan(x) by differentiating both sides of the equivalent equation tan(y)=x.

Accepted Answer

Hello. In this video, we are going to let y equal to cotangent inverse of 2 x + 3. We will find DYD X by differentiating the implicit equation cotangent of Y equal to 2 x + 3. So the goal of this problem is that we're trying to find the implicit derivative DYDX. In order to do this, we will take a look at the given implicit equation. Again, the equation is cotangent of Y equal to 2 x + 3. And in order to find the derivative DYDX, the first thing we want to do is we want to take the derivative of this equation with respect to X. Starting with the left hand side, the derivative of cotangent of y is going to be negative co sequin squared of y. And since this is a derivative of a function of y with respect to x, we will multiply this by the implicit derivative dyy d x. On the right hand side, we will take the derivative term by term, but the derivative of 2 x plus 3 with respect to x is just 2. We need to isolate for the derivative DYDX, so we will divide both sides of this equation by negative cosequin squared of y. That will leave us with Dy D X equal to -2 divided by cosequin squared of y. Now, we need to simplify this into terms of sine and cosine, but recall that cosequent square of Y is equivalent to one divided by sin squared of y. So this is going to simplify to be DYDX equal to -2 divided by 1 divided by sin squared of Y. And by further simplifying this complex fraction, we get the implicit derivative D Y D X equal to -2 multiplied by sin squared of y. Now what we want to do is we want to express sin squad of y in terms of X. In order to do this, let's go ahead and use the original given function and the Pythagorean identity. Now recall that the implicit derivative or the implicit equation given to us is cotangent of y equal to 2 x plus 3. Now, recall to that cotension of Y. Is equivalent to cosine of X, a cosine of y. Divided by sin of y. So this allows us to rewrite the left hand side of the equation as cosine of y divided by sine of Y. Equal to 2 x + 3 And if we were to multiply both sides of this equation by cosine, or by sine to eliminate the denominator, we get cosine of y equal to the quantity 2 x + 3 multiplied by sin of y. Now here, the recall that the Pythagorean identity is sin square of y plus cosine square of y equal to 1. Now, what we can go ahead and do is we can apply this identity of cosine of y into the Pythagorean identity. This will allow us to rewrite the equation of the Pythagorean identity. As sin square of y plus the quantity, 2 x + 3 multiplied by sin of y quantity squared equal to 1. Further simplifying this will leave us with sins squared of y plus. 2 x + 3 quantity squared multiplied by sin squared of y equal to 1. Notice that we can factor out a sin squared of y from the left hand side. If we do this, we get sin squared of y multiplied by the quantity 1 + 2 x + 3 quantity squared equal to 1. And finally, if we divide both sides of this equation by 1 + 2 X + 3 quantity squad, we get sin squared of Y equal to 1 divided by 1 plus the quantity, 2 x + 3 quantity squared. That was a lot of work, but the important thing here is that now we have a substitution for sin squared of y in our implicit differ in our implicit derivative. So by applying this substitution, we get DYDX. Equal to -2 multiplied by the quantity 1 divided by 1 plus the quantity 2 x + 3 quantity squared and finally if we multiply the coefficient of -2 into this quantity, the final derivative is going to bed y d x equal to -2 divided by the quantity 1 + 2 x + 3 squared. This is going to be the implicit derivative that we are looking for. And with that being said, the overall solution to this problem is going to be option C. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

128. Derive the formula dy/dx = 1/(1+x²) for the derivative of y = arctan(x) by differentiating both sides of the equivalent equation tan(y)=x.

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