1. Find (dz)/(dt) for each of the following by using chain...

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Transcript

00:01 Hello everyone in this question.

00:02 We have to find the derivative of the given function.

00:05 So they have given y equal to cos 2x to the power 4 minus 6.

00:11 So first we have to find the derivative that is y dash y dash can be written as minus sign 2x to the power 4 minus 6 then again differentiate 2x to the power 4 minus 6.

00:26 So we will be having minus sign 2x to the power 4 minus 6.

00:31 This will become 2 into 4 into x cube minus 0 will be having minus sign 2x to the power 4 minus 6 8x cube.

00:44 So which can be written as minus 8x cube sine 2x to the power 4 minus 6.

00:54 So from the given option, we can say that option b is correct.

01:01 Then we have to find the second part.

01:09 So the second part is we have to find d square y by dx square.

01:15 The equation is 5 minus 2xy 7x minus 3y.

01:19 So first we can write this as 5 minus 2xy minus 7x plus 3y equal to 0.

01:31 So we keep y as one side x as another side.

01:35 So what we will do means we keep dy minus 2xy to this side and move 5 and 7x to that side.

01:42 So minus 5 plus 7x in we take y as common.

01:47 So we will be having 3 minus 2x minus 5 plus 7x.

01:52 So we can write y as minus 5 plus 7x divided by 3 minus 2x.

02:01 So this is the y...

Question

1. Find $\frac{dz}{dt}$ for each of the following by using chain rule: (a) $z = x^2y$, $x = -t^2$, and $y = 2t^3$ (b) $z = \ln(x + 3y^2)^2$, $x = \sqrt{t}$, and $y = t^{1/4}$.

Question

1. Find $\frac{dz}{dt}$ for each of the following by using chain rule: (a) $z = x^2y$, $x = -t^2$, and $y = 2t^3$ (b) $z = \ln(x + 3y^2)^2$, $x = \sqrt{t}$, and $y = t^{1/4}$.

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