Using the definition of derivative, the limit is H approaches 0 sqrt 11 the quantity Theta plus H -, sqrt 11 Theta all over H First thing we're going to do is multiply by its conjugate in order to get the square roots to go away in the numerator. If I distribute out the 11:00, I'm going to end up with 11 Theta plus eleven H -, 11 Theta. So the 11 thetas are going to cancel and we get eleven H / H times that sqrt 11 Theta plus H plus squared of 11 Theta. So the H is cancel and then when H goes to zero we end up with 11 / sqrt 11 Theta plus squared of 11 Theta. If I have one of them plus another one of them, I have two of them going to rationalize it, get rid of the square root in the denominator. So multiplied by sqrt 11 Theta over sqrt 11 Theta. So we get 11 sqrt 11 Theta over 2 * 11 Theta. The elevens cancel. So here's our final answer for finding the derivative. Then it asks us to find them at specific points. So now all I'm going to do is stick in the values P of one. So every time I see a Theta, I'm going to put in one and I get sqrt 11 / 2 P prime of 11. Every time I see Theta, I'm going to put in 11. Sqrt 11 * 11 is just plain 11 / 2 * 11. So the 11th canceled, leaving us a half P prime of 10 elevenths. Once again, when I see Theta, I'm going to stick in 10 elevenths. The elevenths here I'll cancel, leaving sqrt 10. In the bottom. We end up with 2 * 10 or 20 elevenths. This is a complex fraction, so we're going to take that sqrt 10 and divide it by 20 elevenths. And if we divide by 20 elevenths, remember that's the same thing as multiplying by its reciprocal of 11 twentieths. So 11 sqrt 10 all over 20.