Fractional powers differentiation | Derivative rules | AP Calculus AB | Khan Academy

So we have ( H(x) ) is equal to ( 5x^{1/4} + 7 ) and we want to find what is ( H' ) of 16, or what is the derivative of this function when ( x ) is equal to 16.

And like always, pause this video and see if you can figure it out on your own.

All right, well let's just take the derivative of both sides of this.

On the left-hand side, I'm going to have ( H'(x) ) and on the right-hand side, well, the derivative of the right-hand side, I can just take the derivative of ( 5x^{1/4} ) and add that to the derivative with respect to ( x ) of 7.

So the derivative of ( 5x^{1/4} ) well, I can just apply the power rule here.

You might say, "Wait, wait, there's a fractional exponent," and I would just say, "Well that's okay, the power rule is very powerful."

So we can multiply ( \frac{1}{4} ) times the coefficient, so you have ( 5 \cdot \frac{1}{4} x^{1/4 - 1} ).

That's the derivative of ( 5x^{1/4} ), and then we have plus 7.

Now, what's the derivative of 7 with respect to ( x )?

Well, seven doesn't change with respect to ( x ); the derivative of a constant, we've seen this multiple times, is just zero.

So it's just plus 0.

And now we just have to simplify this, so this is going to be ( H'(x) ) is equal to ( \frac{5}{4} x^{-3/4} + 0 ).

So we don't have to write that.

And now, let's see if we can evaluate this when ( x ) is equal to 16.

So ( H'(16) ) is ( \frac{5}{4} \cdot 16^{-3/4} ).

Well, that's the same thing as ( \frac{5}{4} \cdot \frac{1}{16^{3/4}} ), which is the same thing as ( \frac{5}{4} \cdot \frac{1}{(16^{1/4})^3} ).

And so what is this?

( 16^{1/4} ) is 2, and then you cube that.

2 to the 3 power is 8.

So that's 8, so you have ( \frac{5}{4} \cdot \frac{1}{8} ), which is going to be equal to ( \frac{5 \cdot 1}{4 \cdot 8} ).

And then ( 4 \cdot 8 ) is 32, and we are done.