Question

How do you differentiate `12sqrtx+12^(3sqrtx)+12^(4sqrtx)`?

Answer 1

`y^' = (12 + 3 * 12^(3sqrt(x)) * ln(12) + 4 * 12^(4sqrt(x)) * ln(12))/(2sqrt(x))`

Explanation:

All you really need to use in order to differentiate this function is the chain rule for a constant raised to a variable power.

Now, for a constant `a` raised to avariable power `x`, you have

`color(blue)(a^x = e^(x * ln(a)))`

This means that you can write, using th chain function for `e^u`, with `u = x * ln(a)`

`d/dx(a^x) = d/(du)(e^u) * d/dx(u)`

`d/dx(a^x) = underbrace(e^(x * ln(a)))_(color(red)(=a^x)) * d/dx(x * ln(a))`

`color(blue)(d/dx(a^x) = a^x * ln(a))`

Your function can be rewritten as

`y = 12 * x^(1/2) + 12^(3 * x^(1/2)) + 12^(4 * x^(1/2))`

The derivative of `y` will be

`d/dx(y) = [d/dx(12x^(1/2))] + [d/dx12^(3 * x^(1/2))] + d/dx(12^(4 * x^(1/2)))`

As you can see, you're going to have to use the chain rule again for the last two derivatives. More specifically, you'll have `12^u`, with `u = 3x^(1/2)`, and `12^t`, with `t = 4x^(1/2)`.

This wil get you

`d/dx(12^u) = d/(du)12^u * d/dx(u)`

`d/dx(12^u) = 12^u * ln(12) * d/dx(3 * x^(1/2))`

`d/dx(12^(3x^(1/2))) = 12^(3x^(1/2)) * ln(12) * 3/2x^(-1/2)`

and

`d/dx(12^t) = d/(dt)12^t * d/dx(t)`

`d/dx(12^(4x^(1/2))) = 12^(4x^(1/2)) * ln(12) * 2 * x^(-1/2)`

Plug this into your target derivative to get

`y^' = 6 * x^(-1/2) + 12^(3x^(1/2)) * ln(12) * 3/2x^(-1/2) + 12^(4x^(1/2)) * ln(12) * 2 * x^(-1/2)`

You can simplify this by using `x^(-1/2)` as a common factor

`y^' = x^(-1/2) * 1/2 * [12 + 3 * 12^(3x^(1/2)) * ln(12) + 4 * 12^(4x^(1/2)) * ln(12)]`

Finally, you can rewrite this as

`y^' = color(green)((12 + 3 * 12^(3sqrt(x)) * ln(12) + 4 * 12^(4sqrt(x)) * ln(12))/(2sqrt(x)))`

Answer 2

For `y = 12^sqrtx + 12^root(3)x + 12^root(4)x`, I get: `y' = ln12/12[12^sqrtx 6/sqrtx + 12^root(3)x 4/root(3)x^2 + 12^root(4)x 3/root(4)x^3]`

Explanation:

I wonder if the intended question was to differentiate:

`y = 12^sqrtx + 12^root(3)x + 12^root(4)x`

Which uses the same ideas as the other answer, but the exponents are different:

`y = 12^sqrtx + 12^root(3)x + 12^root(4)x = 12^(x^(1/2)) + 12^(x^(1/3)) + 12^(x^(1/4))`

And the derivative will be:

`y' = 12^(x^(1/2))ln12 d/dx(x^(1/2)) + 12^(x^(1/3))ln12 d/dx(x^(1/3)) + 12^(x^(1/4))ln12 d/dx(x^(1/4))`

`= 12^(x^(1/2))(ln12) [1/2 x^(-1/2)] + 12^(x^(1/3))(ln12) [1/3 x^(-2/3)] + 12^(x^(1/4))(ln12) [1/4 x^(-3/4)]`

`= ln12/12 [ 12^(x^(1/2))6 x^(-1/2) + 12^(x^(1/3)) 4x^(-2/3) + 12^(x^(1/4)) 3 x^(-3/4)`

`= ln12/12[12^sqrtx 6/sqrtx + 12^root(3)x 4/root(3)x^2 + 12^root(4)x 3/root(4)x^3]`