Question

4 If \(y = \left(\frac{x^3 - 2}{2x^5 - 1}\right)^4\), find \(\frac{dy}{dx}\) at \(x = 1\). Select one: a. -52 b. 13 c. -13 d. -28

          4
If \(y = \left(\frac{x^3 - 2}{2x^5 - 1}\right)^4\), find \(\frac{dy}{dx}\) at \(x = 1\).
Select one:
a. -52
b. 13
c. -13
d. -28
        
4
If y = ((x^3 - 2)/(2x^5 - 1))^4, find (dy)/(dx) at x = 1.
Select one:
a. -52
b. 13
c. -13
d. -28

Added by Toni C.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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If y=((x^(3)-2)/(2x^(5)-1))^(4), find (dy)/(dx) at x=1. Select one: a. -52 b. 13 c. -13 d. -28 find dy at x = 1. dx Select one: 0 a. -52 b. 13 c.-13 d.-28
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Transcript

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00:01 In this problem we are provided with the equation negative 2 x squared y minus 7 times x equals to 3y squared plus 2.
00:12 We are also provided that d x over d t equals to 8 x equals to negative 1, y equals to 1 and we are asked to find out the value of d y over d t.
00:25 So let us consider the given equation which is negative 2 x squared y minus 7x equals to 3y squared plus 2.
00:35 And now let us differentiate this with respect to t.
00:39 So in the first term we have the product of two functions.
00:42 So we make use of the product rule of differentiation which is uv dash plus v u dash.
00:48 So here we have negative 2 times the derivative of x squared which is 2 times x times dx over d t since we are differentiating with respect to t times y minus two times x squared times the derivative of y with respect to t which is d y over d t next we have minus 7x which on differentiating with t with respect to t would give us negative 7 times d x over d t this equals to the derivative of 3 y squared which is 3 times 2 y times dy over dt plus the derivative of any constant is 0.
01:33 So now here let us substitute the given values.
01:37 We have negative 2 times 2 times x which is negative 1 times dx over dt which is 8 times y which is 1 minus 2 times x squared which is negative 1 and that is 1 times dy over dt which is 1 minus 2 times x squared...
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