10.
We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here, f(x) = 2 and x = 9. This means that f(9) = 2 = 6 and f(9 +
h) = 2 . If we now plug these into the definition of the derivative, we get f′(9) = = . Notice that if we now take the limit, we get theindeterminate form . With polynomials, we merely simplify the expression to eliminate thisproblem. In any derivative of a square root, we first multiply the top and the bottom of theexpression by the conjugate of the numerator and then we can simplify. Here, the conjugate is . We get f′(9) = × . This simplifies to f′(9) =
= = . Now we can cancel the h in thenumerator and the denominator to get f′(9) = . Now we take the limit: f′(9) = = = .
11.
We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = 5 and x = 8. This means that f(8) = 5 = 20 and f(8
+ h) = . If we now plug these into the definition of the derivative, weget f′(8) = = . Notice that if we now take the limit,we get the indeterminate form . With polynomials, we merely simplify the expression toeliminate this problem. In any derivative of a square root, we first multiply the top and thebottom of the expression by the conjugate of the numerator and then we simplify. Here theconjugate is + 20. We get f′(8) = × . Thissimplifies to f′(8) = = = . Now we