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

##### 2 Answers

#### 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

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

This means that you can write, using th chain function for

#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

#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

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

#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)))#

For

#### Explanation:

I wonder if the intended question was to differentiate:

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

And the derivative will be:

# = 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]#