How do you find the instantaneous rate of change of #2/(sqrt(x) + 1)# using the #lim_(h->0)# method?
1 Answer
See explanation for method.
Explanation:
The limit definition of the derivative states that the derivative of a function
#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#
So here, where
#f'(x)=lim_(hrarr0)(2/(sqrt(x+h)+1)-2/(sqrtx+1))/h#
Now, just power through some algebra:
#f'(x)=2lim_(hrarr0)((sqrtx+1-(sqrt(x+h)+1))/((sqrt(x+h)+1)(sqrtx+1)))/h#
#f'(x)=2lim_(hrarr0)(sqrtx-sqrt(x+h))/(h(sqrt(x+h)+1)(sqrtx+1))#
Multiply the numerator and denominator by the conjugate of the numerator:
#f'(x)=2lim_(hrarr0)((sqrtx-sqrt(x+h))(sqrtx+sqrt(x+h)))/(h(sqrt(x+h)+1)(sqrtx+1)(sqrtx+sqrt(x+h)))#
#f'(x)=2lim_(hrarr0)(x-(x+h))/(h(sqrt(x+h)+1)(sqrtx+1)(sqrtx+sqrt(x+h)))#
#f'(x)=2lim_(hrarr0)(-h)/(h(sqrt(x+h)+1)(sqrtx+1)(sqrtx+sqrt(x+h)))#
#f'(x)=2lim_(hrarr0)(-1)/((sqrt(x+h)+1)(sqrtx+1)(sqrtx+sqrt(x+h)))#
Now we can evaluate the limit, since
#f'(x)=(-2)/((sqrt(x+0)+1)(sqrtx+1)(sqrtx+sqrt(x+0))#
#f'(x)=(-2)/((sqrtx+1)(sqrtx+1)(sqrtx+sqrtx)#
#f'(x)=(-1)/(sqrtx(sqrtx+1)^2)#