To find d
dx f (x), we need to take the derivative of the given function f (x). Using the chain rule, we have:
f (x) = x
2
sin(2x
2 + 20)
f '(x) = 2x sin(2x
2 + 20)⋅ d
dx (x
2) + x
2 ⋅ cos(2x
2 + 20)⋅ d
dx (2x
2 + 20)
f '(x) = 2x sin(2x
2 + 20)⋅ 2x + x
2 ⋅ cos(2x
2 + 20)⋅ 4x
f '(x) = 4x
3 sin(2x
2 + 20) + 4x
3 cos(2x
2 + 20)
Therefore, d
dx f (x) = 4x
3 sin(2x
2 + 20) + 4x
3 cos(2x
2 + 20).
Find d
dx f (x) given
(a) f (x) = x
2
sin(2x
2 + 20)
3 answers
I'm going to assume that they meant
f (x) = x^2 sin(2x^2 + 20)
that makes
df/dx = 2x sin(2x^2 + 20) + 4x^3 cos(2x^2 + 20)
f (x) = x^2 sin(2x^2 + 20)
that makes
df/dx = 2x sin(2x^2 + 20) + 4x^3 cos(2x^2 + 20)
Yes, that is correct. Using the product rule and the chain rule, the correct derivative of f(x) = x^2 sin(2x^2 + 20) with respect to x is:
f'(x) = 2x sin(2x^2 + 20) + x^2 cos(2x^2 + 20) d/dx(2x^2 + 20)
f'(x) = 2x sin(2x^2 + 20) + x^2 cos(2x^2 + 20) (4x)
f'(x) = 2x sin(2x^2 + 20) + 4x^3 cos(2x^2 + 20)
This is the same as the answer you provided.
f'(x) = 2x sin(2x^2 + 20) + x^2 cos(2x^2 + 20) d/dx(2x^2 + 20)
f'(x) = 2x sin(2x^2 + 20) + x^2 cos(2x^2 + 20) (4x)
f'(x) = 2x sin(2x^2 + 20) + 4x^3 cos(2x^2 + 20)
This is the same as the answer you provided.