derivative of;
((1+x^2)^2)sin(2x)
use the product rule
i got (1+x^2)^2(2cos(2x)+sin(2x)(4x(1+x^2)
is that right?
yes, correct
you might want to consider factoring it a bit more
(I see 2(1+x^2) as a common factor)
thanks you so much. that really helped alot
To find the derivative of the given function, which is ((1+x^2)^2)sin(2x), we can use the product rule and chain rule of derivatives. Here's how you can do it step by step:
Step 1: Apply the product rule.
The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is given by:
(d/dx) [u(x) * v(x)] = u(x) * v'(x) + v(x) * u'(x),
where u'(x) and v'(x) represent the derivatives of u(x) and v(x), respectively.
Let's assign u(x) = (1+x^2)^2 and v(x) = sin(2x).
Step 2: Find the derivative of u(x).
To differentiate u(x) = (1+x^2)^2, we can use the chain rule.
Let's define w(x) = 1+x^2. Then the derivative of w(x)^2 is given by:
(d/dx) [w(x)^2] = 2w(x) * w'(x).
Now let's find w'(x):
w(x) = 1+x^2,
w'(x) = 2x.
Substituting these values back into the derivative of u(x), we get:
(d/dx) [u(x)] = 2(1+x^2) * 2x.
Simplifying this expression, we have:
(d/dx) [u(x)] = 4x(1+x^2) = 4x + 4x^3.
Step 3: Find the derivative of v(x).
To find the derivative of v(x) = sin(2x), we can use the chain rule.
Let's define z(x) = 2x. Then the derivative of sin(z(x)) is given by:
(d/dx) [sin(z(x))] = cos(z(x)) * z'(x).
Now let's find z'(x):
z(x) = 2x,
z'(x) = 2.
Substituting these values back into the derivative of v(x), we have:
(d/dx) [v(x)] = cos(2x) * 2.
Simplifying this expression, we get:
(d/dx) [v(x)] = 2cos(2x).
Step 4: Apply the product rule.
Using the product rule, the derivative of the original function can be calculated as follows:
(d/dx) [((1+x^2)^2)sin(2x)] = u(x) * v'(x) + v(x) * u'(x).
Substituting the values we found earlier, we have:
(d/dx) [((1+x^2)^2)sin(2x)] = (1+x^2)^2 * 2cos(2x) + sin(2x) * (4x + 4x^3).
Expanding and simplifying this expression will give you the derivative of the function.