# Find y’ and y”.y = sin(x^2)?

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Find y’ and y”.y = sin(x^2)?

• y = sin(x^2)

y’ = 2x*cos(x^2)

y” = -4x^2*sin(x^2) + 2*cos(x^2)

• You need to know your TRIG FUNCTIONS for the first and second derivatives:

y’ = -sin(x^2)(2x)

y’ = -2xsin(x^2)

For the second derivative, you need to know the PRODUCT RULE:

y” = (-2x)*(cos(x^2))(2) + (sin(x^2))*(-2)

Almost everyone who answered this question did not do the second derivative correctly from trig functions. REMEMBER: derivative of cosx = -sinx and derivative of sinx = cosx; the -2 found is considered with the variable it’s attached to, which is x (i.e. -2x)

y” = -4xcos(x^2) -2sin(x^2)

• y’ = 2xcos(x²)

y” = -4x²sin(x²)

Edit: uh, oops. Forgot about that pesky x from the first derivative and needing to do the product rule. My math professors would all bonk me on the head for that one.

y” should be -4x²sin(x²) + 2cos(x²)

• For this you must use the chain rule. u = x^2 du/dx = 2x

dsin(u)/dx = 2x(cos(x^2)) = y’

Use the product rule to find d2xcos(x^2)/dx

Factor out constants. 2(dxcos(x^2)/dx)

Product rule is: d(uv)/dx=v(du/dx)+u(du/dx) where u=x v=cos(x^2)

Derive cos(x^2) with chain rule, where u = x^2 and du/dx = 2x.

d(uv)/dx = 2(cos(x^2)(1))+(x(-2xsin(x^2)) = 2(cos(x^2))-((2x^2)(sin(x^2)))

Thus, y’= 2x(cos(x^2))

and y”= 2 (cos(x^2)-x (2 x) sin(x^2))

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