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y = (x^2)(sin x)(2x)(cos x) - 2sin x
y' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'
y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'
y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)
y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)
y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos x
y' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)
y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos x
y' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos x
y' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
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What is the derivative of -2sin x?
The derivative of sin(x) is cos(x). Coefficients act like constants and always remain in derivatives. So, the derivative is -2cos(x).
What is the 87th derivative of sin x?
Every fourth derivative, you get back to "sin x" - in other words, the 84th derivative of "sin x" is also "sin x". From there, you need to take the derivative 3 more times, getting:85th derivative: cos x86th derivative: -sin x87th derivative: -cos x
What is the Derivative of sin?
Generally, the derivative of sine is cosine.
How do you find the derivative of y equals sin8x?
Derivative of sin x = cos x, so chain rule to derive 8x = 8 , answer is 8cos8x
What is the derivative of root sin2x?
You are supposed to use the chain rule for this. First step: derivative of root of sin2x is (1 / (2 root of sin 2x)) times the derivative of sin 2x. Second step: derivative of sin 2x is cos 2x times the derivative of 2x. Third step: derivative of 2x is 2. Finally, you need to multiply all the parts together.
