What is the derivative of (x^4/3)+4sinx-(2/x^2)?

Question

What is the derivative of y = [(e^-x) - (e^x)]^3

Answer 1

So we want f'(x) if

f(x) = (x^4)/3 + 4sin(x) - (2/(x^2))

One thing we can do is change 2/x^2 into 2x^(-2). We can also pull out the constant (1/3) out of the first term.

f(x) = (1/3) x^4 + 4sin(x) - 2x^(-2)

Now, we differentiate.

f'(x) = (1/3) [4x^3] + 4cos(x) + 4x^(-3)

Cleaning this up a bit,

f'(x) = (4/3)x^3 + 4cos(x) + 4/(x^3)

Answer 2

With addition, the derivative of the whole solution is the sum of the derivatives of its parts.

So, first, use the power rule. 4/3x^(1/3) Turn the x^(1/3) into a radical (the cubed root of x).

Next, take the derivative of 4sinx. The derivative of a sine is always cosine, so the derivative of 4sinx is 4cosx.

Lastly, take the derivative of the last part, which is -2/(x^2). To simplify this, make it -2*x^-2. This will be 4x^-3. This is the same as 4/(x^3).

Now, add them all together, getting
4/(3*the cubed root of x)+4cosx+4/(x^3)

On the second one you will need the chain rule. First, find dy/du Let u=(e^-x)-(e^x).

dy/du will be 3u^2

now find du/dx. Take the derivative of u.

e^/-x is just 1/(e^x), so the derivative of that is (-e^x)/(e^2x)

The derivative of e^x is just e^x, so the derivative of u is

(-e^x)/(e^2x)-e^x.

Now, we must multiply dy/du and du/dx to get dy/dx.

3u^2 x (-e^x)/(e^2x)-e^x

We must substitute u, so

3((e^-x)-(e^x))^2 x (-e^x)/(e^2x)-e^x

This is your answer.

I hope that this helped.

Answer 3

4/3(x^1/3) + 4cosx - 4x^-3
3[(e^-x)-(e^x)]^2 [(e^-x)-(e^x)]

Answer 4

(4/3)x^3+4cosx+(4/x^3)

2.3[e^-x-e^x]^2(-e^-x-e^x)