Second Derivatives

In Exerci...

Question

Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

3 + sin y = y – x³

Accepted Answer

Welcome back, everyone. Using implicit differentiation, determine the correct expression for the 1st and 2nd derivatives. 4 plus cosine of Y equals Y minus X2. So let's begin with the first derivative. What we're going to do is simply differentiate both sides of the given equation. Let's valuate the divided by the X of 4 plus cosine of Y. And equate it to D divided by D X. Of Y minus X squared. Now we're going to identify each derivative, so the derivative of 4 is 0. The derivative of cosine is negative sign, so we get negative sign of Y. But Y is a function of X, so we are applying the chain rule and multiplying by D Y divided by D X. This is going to be equal to the derivative of Y, which is D Y divided by D X. Minus the derivative of X squad which is 2 X. So now what we're going to do is simply move D Y divided by DX to the left, which gives us negative D Y. Divided by D X minus. D Y divided by DX multiplied by sin of Y, and this is equal to -2 X. Now we can remove all of the negative signs by basically multiplying our equation by -1. So all of the signs become positive and now we can factor out D Y divided by D X. In priss we get one. Plus sign of Y. And this is equal to 2 X. So now solving for the first derivative, we can show that D Y divided by DX is the ratio between 2 X and the factor adjacent to Dy divided by DX. That's sin of Y + 1. Well done. So we have our first answer. And now we can continue. And find the second derivative. Once again, we can use implicit differentiation, so we're taking the derivative of D Y divided by DX. Which is the second derivative of Y with respect to X. And this is going to be equal to D divided by D X. Off the right hand side. 2 X divided by sin of Y + 1. As we understand, we're going to apply the quotient rule, because we have the irrational expression. And according to the quotient rule, we're taking the bottom function. Sign of Y + 1. Multiplying it by the derivative. Of the top function, which is the derivative of 2 X. And then we're subtracting the top function multiplied by the derivative of the bottom function. So, minus 2 X multiplied by D divided by the X of sin of Y + 1. All of that is going to be divided by the square of sin of Y + 1. Let's identify each derivative in our numerator. The derivative of 2 X is just 2, so we get 2 multiplied by Sign of Y + 1. Minus 2 X. Multiplied by the derivative of sin of Y + 1. The derivative of 1 is 0 because it's a constant. The derivative of sine is cosine, so we get cosine of Y. And because Y is a function of X, we're simply applying. The chain rule, so we're multiplying by the derivative of Y. And the derivative of Y was previously calculated, we had 2 X. Divided by a sign of Y. Plus one. And then we can just rewrite the denominator sign of Y + 1. Square What we can do from here is basically use the common denominator, right? And essentially we can write sin of Y + 1 cubed. Within our fraction. So that in the numerator we can multiply all of our terms by sign of Y plus 1. Which gives us 2. Sign of Y plus 1 square, we're squaring it, right, because we're multiplying sign of Y plus 1 by itself, and then we are subtracting. 2 X cosine off Y. Multiplied by 2 x, so that would be or X2d cosine of y. Notice that since we're multiplying by sin of y plus 1, we're simply getting rid of the denominator. And this essentially provides us with the second derivative of Y. Well then, let's label our final answers to this problem. Thank you for watching.

Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

3 + sin y = y – x³

Key Concepts

Implicit Differentiation

First Derivative (dy/dx)

Second Derivative (d²y/dx²)

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