Question

Question #bf2cc

Answer 1

Use the Power Rule.
`(df)/dx = 4/3*(4x)^(-2/3) + 5/2 x^(-3/2) = 4/(3(4x)^(2/3)) + 5/(2x^(3/2))`

Explanation:

We must use the Power Rule here, which states:

`f(x) = ax^n + bx^(n-1) + ...+ jx + k, (df)/dx = nax^(n-1) + (n-1)bx^(n-2) + ...+ j`

We have the cube root of 4x, minus five divided by the square root of x, minus the natural logarithm of 5.

Thus:

`f(x) = (4x)^(1/3) - 5x^(-1/2) - ln5`

For this first term, we must remember the chain rule, which states that:

`y=f(g(x)), dy/dx = g'(x) * f'(g)`

With `f(g) = g^(1/3), g(x) = 4x`, we have `f'(g) = 1/3 x^(-2/3), g'(x) = 4`

Returning to our initial equation...

`(df)/dx = 4/3*(4x)^(-2/3) + 5/2 x^(-3/2) = 4/(3(4x)^(2/3)) + 5/(2x^(3/2))`