How do you find the derivative of f(x) = -2sqrt(x) by the limit process?

f(x)=#-2sqrt(x)#

2 Answers

Answer:

#f'(x)=-1/\sqrtx#

Explanation:

Given function: #f(x)=-2\sqrtx# then its derivative using first principle as follows

#\frac{d}{dx}f(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}#

#f'(x)=\lim_{h\to 0}\frac{-2\sqrt{x+h}-(-2\sqrtx)}{h}#

#=\lim_{h\to 0}\frac{-2\sqrtx(\sqrt{\frac{x+h}{x}}-1)}{h}#

#=-2\sqrtx\lim_{h\to 0}\frac{\sqrt{(1+h/x)}-1}{h}#

#=-2\sqrtx\lim_{h\to 0}\frac{(1+h/x)^{1/2}-1}{h}#

#=-2\sqrtx\lim_{h\to 0}\frac{(1+(1/2) h/x+(1/2)(1/2-1)(h/x)^2+\ldots)-1}{h}#

#=-2\sqrtx\lim_{h\to 0}\frac{(1/2) h/x+(1/2)(-1/2)(h/x)^2+\ldots}{h}#

#=-2\sqrtx\lim_{h\to 0}((1/2) 1/x+(1/2)(-1/2)h/x^2+\ldots)#

#=-2\sqrtx((1/2) 1/x+0)#

#=-1/\sqrtx#

Jul 7, 2018

By definition:

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

Let: #X = x + h#

#=- 2 lim_(X to x) (sqrtX - sqrtx)/(X - x) * (sqrtX + sqrtx)/(sqrtX + sqrtx)#

#=-2 lim_(X to x) ( X - x)/(X - x) * 1/(sqrtX + sqrtx)#

#=-2 lim_(h to 0) 1/(sqrt(x + h) + sqrtx) = -1/( sqrtx)#