find derivative of sin2x using delta method
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By the chain rule:
(f∘g(x))'=f'(g(x))⋅g'(x)
This means that the derivative of "f of g of x" is equal to the derivative of "f of g of x" with respect to f multiplied by the derivative of "g of x."
We take the derivative of the outermost term first, and then multiply by the derivative of the inside term. We're given the function:
f(x)=sin(2x)
The outermost term is the sine function. We know that the derivative of the sine is the cosine. The innermost term is 2x. The derivative of 2x is simply 2.
f'(x)=cos(2x)⋅2
⇒f'(x)=2cos(2x)
(f∘g(x))'=f'(g(x))⋅g'(x)
This means that the derivative of "f of g of x" is equal to the derivative of "f of g of x" with respect to f multiplied by the derivative of "g of x."
We take the derivative of the outermost term first, and then multiply by the derivative of the inside term. We're given the function:
f(x)=sin(2x)
The outermost term is the sine function. We know that the derivative of the sine is the cosine. The innermost term is 2x. The derivative of 2x is simply 2.
f'(x)=cos(2x)⋅2
⇒f'(x)=2cos(2x)
harry003:
its not delta method
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