Find d
dx f (x) given
(a) f (x) = x
2
sin(2x
2 + 20)
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)
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.
To find d/dx f(x) for f(x) = x^2 sin(2x^2 + 20), we will use the product rule and chain rule of differentiation.
Let's break down the function f(x) into two parts: g(x) = x^2 and h(x) = sin(2x^2 + 20).
Now, let's find the derivative of g(x) first:
g'(x) = d/dx (x^2)
= 2x
Next, let's find the derivative of h(x):
h'(x) = d/dx [sin(2x^2 + 20)]
To differentiate this, we use the chain rule. The chain rule states that if we have a composite function y = f(g(x)), then its derivative is given by dy/dx = f'(g(x)) * g'(x). In this case, our outer function is sin(u), where u = 2x^2 + 20.
So, let's differentiate the outer function, which is sin(u):
d/dx [sin(u)] = cos(u)
Now, let's differentiate the inner function u with respect to x:
du/dx = d/dx (2x^2 + 20)
= 4x
Therefore, applying the chain rule, we get:
h'(x) = cos(u) * du/dx
= cos(2x^2 + 20) * 4x
Finally, we can find d/dx f(x) by using the product rule:
d/dx [f(x)] = g'(x) * h(x) + g(x) * h'(x)
= 2x * sin(2x^2 + 20) + x^2 * cos(2x^2 + 20) * 4x
So, the derivative of f(x) = x^2 sin(2x^2 + 20) with respect to x is:
d/dx [f(x)] = 2x * sin(2x^2 + 20) + 4x^3 * cos(2x^2 + 20)
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).