How would you find the derivative of f(x)=1/sqrtx using limits? I know how to do problems with a fraction or a square root, but I can't figure out how to do one with both. I have tried a few different starts, but always end up getting stuck.
We will be using the binomial expansion
(1+x)n
=1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! + ...
We will be calculating, for f(x)=1/√x, x≠0, h≠0 and let a=h/x
D(x,h)=(1/√(x+h) - 1/√x)/h
=(1/√x)((1/√(1+a)-1/√1))/h
=(1/√x)(1 + (-1/2)a + (-1/2)(-3/2)a²/2! + ... -1)/h
=-(1/√x)((1/2)(h/x) - (3/8)(h/a)² + ...)/h
=-(1/√x)(1/x - (3/8)(h/x²) + ...)
f'(x) = Lim (h→0)(D(x,h))
=-1/√x)
=-(1/√x)(1/x)
=-x-3/2
f'(x) = Lim (h→0)(D(x,h))
=-(1/√x)(1/x)
=-x-3/2
To find the derivative of a function that involves both a fraction and a square root, such as f(x) = 1/√x, we can use the definition of the derivative involving limits. Here's how you can proceed:
Step 1: Begin by expressing the function in a different form that makes it easier to work with. In this case, we can rewrite f(x) as f(x) = x^(-1/2).
Step 2: Now, we can apply the limit definition of the derivative. The derivative of a function f(x) is given by:
f'(x) = lim(h->0) [f(x + h) - f(x)] / h
Step 3: Substitute the expression for f(x) into the limit definition:
f'(x) = lim(h->0) [ (x + h)^(-1/2) - x^(-1/2) ] / h
Step 4: Simplify within the limit expression. To do this, we'll have to use the difference of squares formula:
(a + b)(a - b) = a^2 - b^2
Using the difference of squares formula on (x + h)^(-1/2) - x^(-1/2), we get:
[(x + h)^(-1/2) - x^(-1/2)] * [(x + h)^(-1/2) + x^(-1/2)]
Step 5: Continuing from the previous step, multiply the numerator and denominator of the limit expression by the conjugate [(x + h)^(-1/2) + x^(-1/2)]:
f'(x) = lim(h->0) [ (x + h)^(-1/2) - x^(-1/2) ] * [(x + h)^(-1/2) + x^(-1/2)] / h * [(x + h)^(-1/2) + x^(-1/2)]
Step 6: Simplify the expression further by observing that the conjugate forms a difference of squares:
[(x + h)^(-1/2) - x^(-1/2)] * [(x + h)^(-1/2) + x^(-1/2)] = [(x + h)^(-1/2)]^2 - [x^(-1/2)]^2
= (x + h)^(-1) - x^(-1)
= 1 / (x + h) - 1 / x
Step 7: Rewrite the limit expression using the simplified forms:
f'(x) = lim(h->0) [ 1 / (x + h) - 1 / x ] / h * [(x + h)^(-1/2) + x^(-1/2)]
Step 8: Split the limit into two parts by applying the limit to each fraction individually:
f'(x) = lim(h->0) [1 / (x + h)] / h * [(x + h)^(-1/2) + x^(-1/2)] - lim(h->0) [1 / x ] / h * [(x + h)^(-1/2) + x^(-1/2)]
Step 9: Evaluate the two limits separately. The first limit simplifies as follows:
lim(h->0) [1 / (x + h)] / h = lim(h->0) 1 / [h(x + h)]
Since h is approaching zero, we can cancel it from the numerator and denominator:
lim(h->0) 1 / [h(x + h)] = 1 / x
Similarly, the second limit can be evaluated in a similar manner:
lim(h->0) 1 / x / h = 1 / x
Step 10: Plug the results of the limits back into the original expression:
f'(x) = (1 / x) * [(x + h)^(-1/2) + x^(-1/2)] - (1 / x) * [(x + h)^(-1/2) + x^(-1/2)]
Step 11: Simplify the expression:
f'(x) = (1 / x) * [(x + h)^(-1/2) + x^(-1/2) - (x + h)^(-1/2) - x^(-1/2)]
Step 12: Notice that we can cancel out terms:
f'(x) = (1 / x) * [x^(-1/2) - x^(-1/2)]
Step 13: Simplify further, noting that the terms cancel each other out:
f'(x) = 0
Therefore, the derivative of f(x) = 1/√x is f'(x) = 0.
I hope this step-by-step explanation helps you understand how to find the derivative of a function involving both a fraction and a square root using the limit definition.