If xexy + ye- xy = sin2x, then dydx at x = 0 is f

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751.

If xexy + ye- xysin2x, then dydx at x = 0 is

  • 2y2 - 1

  • 2y

  • y2 - y

  • y2 - 1


D.

y2 - 1

Given, xexy + ye- xy = sin2xOn differentiating w.r.t. x, we getexy + xexyxdydx + y + dydxe- xy - ye- xyxdydx + y = 2sinx . cosx x2exy + e- xy - xeye- xydydx + exy + xyexy - y2e- xy = sin2xOn putting x = 0, we get0 + e- 0 - 0dydxx = 0 + e0 + 0 - y2e- 0 = sin0 1dydxx  =0 + 1 - y2 = 0 dydxx = 0 = y2 - 1


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752.

If y = tan-12x - 11 + x - x2, then dydx at x = 1 is equal to

  • 12

  • 23

  • 1

  • 32


753.

If f(x) = cos-12cosx + 3sinx13, then [f'(x)]2 is equal to

  • 1 + x

  • 1 + 2x

  • 2

  • 1


754.

If u = tan-11 - x2 - 1x and v = sin-1x, then dudv is equal to

  • 1 - x2

  • - 12

  • 1

  • - x


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755.

If y = 11 + x + x2, then dydx is equal to

  • y2(2 + 2x)

  • - 1 + 2xy2

  • 1 + 2xy2

  • - y2(1 + 2x)


756.

If g(x) is the inverse of f(x) and f'(x) = 11 + x3, then g'(x) is equal to

  • g(x)

  • 1 + g(x)

  • 1 + {g(x)}3

  • 11 + gx3


757.

If y = f(x2 + 2) and f'(3) = 5, then dydx at x = 1 is

  • 5

  • 25

  • 15

  • 10


758.

Let, f(x) = x2 + bx + 7. If f'(5) = 2f'72, then the value of b is

  • 4

  • 3

  • - 4

  • - 3


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759.

If y = sin-12x1 - x2, - 12  x  12, then dydx is equal to

  • x1 - x2

  • 11 - x2

  • 21 - x2

  • 2x1 - x2


760.

The function fx = 2x2 - 1,     if 1  x  4151 - 30x, if 4 < x  5 is not suitable to apply Rolle's theorem, since

  • f(x) is not continuous on [1, 5]

  • f(1) f(5)

  • f(x) is continuous only at x = 4

  • f(x) is not differentiable at x = 4


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