FInd the derivative of each of the following function.
f(x)=xcosx+x^4e^2x
All I know is the rule with the e^u equation where you would do e^2x times 1. other than that I dont know what to do can someone please help and show all work I will appreciate it greatly
f' = x d/dx(cos x) + cos x dx/dx +x^4d/dx(e^2x) + e^2x d/dx(x^4)
= -x sin x + cos x + 2x^4(e^2x) + 4x^3 (e^2x)
just the good old chain rule and product rule
d/dx e^u = e^u du/dx
d/dx uv = u'v + uv'
so, with
y = x cosx + x^4 e^2x
y' = (1)cosx + (x)(-sinx) + (4x^3)(e^2x) + (x^4)(e^2x)(2)
To find the derivative of the function f(x) = xcosx + x^4e^2x, we can apply the product rule and the chain rule. Let's go step by step:
1. Product Rule: The product rule allows us to find the derivative of a function that is the product of two or more functions. The formula for the product rule is (u*v)' = u'v + uv'.
Let's label the two parts of f(x) as follows:
u = xcosx
v = x^4e^2x
Now, we can find the derivatives of u and v.
2. Derivative of u: Using the product rule, we need to find the derivatives of x and cosx separately.
u = xcosx
u' = (x)'cosx + x(cosx)' = cosx + x(-sinx) = cosx - xsinx
3. Derivative of v: Here, we also need to apply both the chain rule and the product rule.
v = x^4e^2x
v' = (x^4)'e^2x + x^4(e^2x)' = 4x^3e^2x + x^4(2e^2x) = 4x^3e^2x + 2x^4e^2x
4. Put it all together: Now that we have obtained the derivatives of u and v, we can use the product rule formula.
f'(x) = u'v + uv'
= (cosx - xsinx)(x^4e^2x) + (xcosx)(4x^3e^2x + 2x^4e^2x)
Simplifying this expression further:
f'(x) = (cosx - xsinx)(x^4e^2x) + (4x^4cosx + 2x^5cosx)e^2x
So, the derivative of the function f(x) = xcosx + x^4e^2x is f'(x) = (cosx - xsinx)(x^4e^2x) + (4x^4cosx + 2x^5cosx)e^2x.