Find the derivative for k(x) = (5x4 + 2)(3sin x).
Would this be the answer:
(60sinx)x^3+ (15cosx)x^5+ 2
No.
Use the product rule
d/dx(u*v) = u dv/dx + v du/dx
and let 5x^4 +2 = u(x), and
3 sin x = v(x)
du/dx = 20 x^3
dv/dx = 3 cos x
Derivative = 3 cos x *(5x^4 + 2)
+ 3 sin x *(20 x^3)
That is how I got my answer, by using product rule.
Obviously you made a mislake somewhere along the line. If you care to show your steps, someone will show you where the error was made.
To find the derivative of the given function k(x) = (5x^4 + 2)(3sin x), we can use the product rule.
The product rule states that if u(x) and v(x) are two differentiable functions, then the derivative of their product is given by:
(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)
Let's apply the product rule to find the derivative of k(x).
Given:
u(x) = 5x^4 + 2
v(x) = 3sin(x)
First, find the derivative of u(x):
u'(x) = d/dx (5x^4 + 2)
= 20x^3
Next, find the derivative of v(x):
v'(x) = d/dx (3sin(x))
= 3cos(x)
Now, apply the product rule:
(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)
= (20x^3)(3sin(x)) + (5x^4 + 2)(3cos(x))
Simplifying this expression gives the derivative of k(x):
k'(x) = (60x^3sin(x)) + (15x^4 + 6cos(x))
So, the correct answer for the derivative of k(x) is:
k'(x) = 60x^3sin(x) + 15x^4 + 6cos(x)