How do you find the derivative of #f(x)=4/sqrtx# using the limit process?

Question

Ezra Adams · Accepted Answer

#f'(x)=(-2)/(xsqrtx)#

The derivative of the quotient is determined by using the definition of derivatives:

#color(red)(f'(x)=lim_(h->0)(f(x+h)-f(x))/h)#

In this exercise we will also apply this polynomial identity to expand: #color(brown)((a^2-b^2)=(a-b)(a+b))#

#f'(x)=lim_(h->0)(4/(sqrt(x+h))-4/sqrtx)/h#

#f'(x)=lim_(h->0)((4sqrtx-4sqrt(x+h))/(sqrt(x+h)sqrtx))/h#

#f'(x)=lim_(h->0)((4sqrtx-4sqrt(x+h)))/(h(sqrt(x+h)sqrtx)#

To get rid of the square root in the numerator we will multiply by its conjugate #color(blue)(4sqrtx+4sqrt(x+h))#

#f'(x)=lim_(h->0)((4sqrtx-4sqrt(x+h))(color(blue)(4sqrtx+4sqrt(x+h))))/(h(sqrt(x+h)sqrtx)(color(blue)(4sqrtx+4sqrt(x+h))))#

#f'(x)=lim_(h->0)((4sqrtx)^2-(4sqrt(x+h))^2)/(h(sqrt(x+h)sqrtx)(4sqrtx+4sqrt(x+h)))#

#f'(x)=lim_(h->0)(16x-16(x+h))/(h(sqrt(x+h)sqrtx)(4sqrtx+4sqrt(x+h)))#

#f'(x)=lim_(h->0)(cancel(16x)cancel(-16x)-16h)/(h(sqrt(x+h)sqrtx)(4sqrtx+4sqrt(x+h)))# #f'(x)=lim_(h->0)(-16cancel(h))/(cancelh(sqrt(x+h)sqrtx)(4sqrtx+4sqrt(x+h)))#

Now let us find the limit #color(green)(lim_(h->0))#

#f'(x)=lim_(h->0)(-16)/((sqrt(x+h)sqrtx)(4sqrtx+4sqrt(x+h)))#

#f'(x)=(-16)/((sqrt(x+0)sqrtx)(4sqrtx+4sqrt(x+0)))#

#f'(x)=(-16)/((sqrt(x)sqrtx)(4sqrtx+4sqrt(x)))#

#f'(x)=(-cancel16)/(x(cancel8sqrtx))#

#f'(x)=(-2)/(xsqrtx)#

Isaiah Allen · Answer

undefined To find the derivative of $ f(x) = \frac{4}{\sqrt{x}} $ using the limit process, you can follow these steps:

1. Begin with the definition of the derivative: $ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $.
2. Substitute the function $ f(x) = \frac{4}{\sqrt{x}} $ into the definition.
3. Simplify the expression $ f(x + h) - f(x) $.
4. Divide the result by $ h $.
5. Take the limit as $ h $ approaches 0.

By following these steps, you will obtain the derivative of $ f(x) = \frac{4}{\sqrt{x}} $ using the limit process.