If tan?¹ y = x² + y², then y' is Select one: a. 2x(1 + y²) /...

If tan?¹ y = x² + y², then y' is Select one: a. 2x(1 + y²) / (1 + 2y + 2y³) b. (1 - 2y - 2y³) / (2x(1 + y²)) c. (1 + 2y + 2y³) / (2x(1 + y²)) d. 2x(1 + y²) / (1 - 2y - 2y³) The derivative of the function ln(x?) - (ln x)? is equal to Select one: a. 8/x (1 - (ln x)?) b. 8 ln x? - 8(ln x)? c. 8/x ln x? - 8(ln x)? d. 8x ln x? - 8/x (ln x)?

Transcript

00:01 Hi there in this question.

00:01 We have two sub -questions.

00:03 The first one that it's asked to find y -dash where tan inverse y is equal to x square plus y square.

00:09 And second one, we have to find the derivative of lnx parate minus lnx whole power rate.

00:15 So here, let's see how we'll do this.

00:18 So for doing the first one, we'll do the differentiation on both the sides of the equation.

00:24 So that we'll be getting d over d x of tan inverse y is equal to d over d.

00:30 Dx of x square plus y square.

00:35 So from here we'll be getting this is equal to from here, this part we'll be getting this implies that we'll be having tan inverse y.

00:43 The derivative of tan inverse y is 1 plus y square.

00:46 Multiplied by d .y over d x because y is considered as a function of x.

00:51 Then we'll be getting this is equal to.

00:56 Now we'll be differentiating the thing here that is 2x plus 2y, dy over d x.

01:01 So what i'm going to do is that i'm going to take the terms involving dy y over d x to the left part.

01:08 So i'll be having 1 divided by 1 plus y square y over d x minus 2y dy over d x is equal to 2x.

01:18 So this implies d y over d x multiplied by 1 divided by 1 plus y square minus 2y is equal to 2x.

01:35 This implies, dy over dx multiplied by 1 minus 2y multiplied by 1 plus y square, whole divided by 1 plus y square is equal to 2x.

01:49 So again we'll be getting that...

Question

Please give Ace some feedback