Question #6f00a

1 Answer
Jul 31, 2015

#y^' = -1/2 * x^(-3/2)#

Explanation:

In order to differentiate this function from the first principle, you need to use the formula

#color(blue)(d/dx(y) = lim_(h->0)(f(x+h) - f(x))/h#

In your case, you will have

#d/dx(y) = lim_(h->0)(1/sqrt(x+h) - 1/sqrt(x))/h#

#d/dx(y) = lim_(h->0) (sqrt(x) - sqrt(x+h))/(h * sqrt(x) * sqrt(x+h))#

Multiply this expression by

#1 = (sqrt(x) + sqrt(x+h))/(sqrt(x) + sqrt(x+h))#

to get

#d/dx(y) = lim_(h->0)[(sqrt(x) - sqrt(x+h))(sqrt(x) + sqrt(x+h))]/(h * sqrt(x) * sqrt(x+h) * (sqrt(x) + sqrt(x+h))#

Use the fact that

#color(blue)( (a-b)(a+b) = a^2 - b^2)#

to simplify this expression

#d/dx(y) = lim_(h->0) ( (sqrt(x))^2 - (sqrt(x+h))^2)/((h * sqrt(x) * sqrt*x) * sqrt(x+h) + h * sqrt(x) * sqrt(x+h) * sqrt(x+h))#

#d/dx(y) = lim_(h->0)(color(red)(cancel(color(black)(x))) - color(red)(cancel(color(black)(x))) -h)/(h * x * sqrt(x+h) + h * (x+h) * sqrt(x))#

#d/dx(y) = lim_(h->0)(-color(blue)(cancel(color(black)(h))))/(color(blue)(cancel(color(black)(h))) * x * sqrt(x+h) + color(blue)(cancel(color(black)(h))) * (x+h) * sqrt(x))#

#d/dx(y) = -1/(x * sqrt(x + 0) + (x + 0) * sqrt(x))#

#d/dx(y) = -1/(2xsqrt(x)) = -1/2 * 1/sqrt(x^3) = color(green)(-1/2 * x^(-3/2))#