How do you use the definition of a derivative to find the derivative of #(2/sqrt x)#?
1 Answer
Explanation:
The limit definition of a derivative states that for a function
#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#
Here,
#f'(x)=lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h#
Multiply the entire fraction by
#f'(x)=lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h((sqrt(x+h)sqrtx)/(sqrt(x+h)sqrtx))#
#f'(x)=lim_(hrarr0)(2sqrtx-2sqrt(x+h))/(hsqrt(x+h)sqrtx)#
#f'(x)=lim_(hrarr0)(2(sqrtx-sqrt(x+h)))/(hsqrt(x+h)sqrtx)#
Multiply by the conjugate of the term in the numerator.
#f'(x)=lim_(hrarr0)(2(sqrtx-sqrt(x+h)))/(hsqrt(x+h)sqrtx)((sqrtx+sqrt(x+h))/(sqrtx+sqrt(x+h)))#
#f'(x)=lim_(hrarr0)(2(x-(x-h)))/(hsqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))#
#f'(x)=lim_(hrarr0)(2h)/(hsqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))#
#f'(x)=lim_(hrarr0)2/(sqrt(x+h)sqrtx(sqrtx+sqrt(x+h)))#
Evaluate the limit by plugging in
#f'(x)=2/(sqrt(x+0)sqrtx(sqrtx+sqrt(x+0)))#
#f'(x)=2/(sqrtxsqrtx(sqrtx+sqrtx)#
#f'(x)=2/(x(2sqrtx))#
#f'(x)=1/(xsqrtx)=1/x^(3/2)#