# How to use the first principle in finding derivative of f(x)=square root (1-x)?

justaguide | Certified Educator

Using the first principles the derivative of f(x) = sqrt (1 - x) is given as lim h-->0[(f(x + h) - f(x))/h]

lim h-->0[(f(x + h) - f(x))/h]

=> lim h-->0[(sqrt(1 - x - h) - sqrt (1 - x))/h]

=> lim h-->0[(sqrt(1 - x - h) - sqrt (1 - x))*(sqrt(1 - x - h) + sqrt (1 - x))/h*(sqrt(1 - x - h) + sqrt (1 - x))]

=> lim h-->0[(1 - x - h - 1 + x)/h*(sqrt(1 - x - h) + sqrt (1 - x))]

=> lim h-->0[(-h)/h*(sqrt(1 - x - h) + sqrt (1 - x))]

=> lim h-->0[-1/(sqrt(1 - x - h) + sqrt (1 - x))]

substitute h = 0

=> (-1/2)*(1/sqrt (1 - x))

The required derivative is (-1/2)*(1/sqrt (1 - x))

giorgiana1976 | Student

We'll recall the first principle:

lim [f(x+h) - f(x)]/h, for h->0

Comparing, we'll get:

lim {sqrt [1 - (x+h)] - sqrt(1-x)}/h

We'll remove the brackets at radicand:

lim [sqrt (1 - x - h) - sqrt(1 - x)]/h

We'll multiply both, numerator and denominator, by the conjugate of numerator:

lim [sqrt (1 - x - h) - sqrt(1 - x)]*[sqrt (1 - x - h) + sqrt(1 - x)]/h*[sqrt (1 - x - h) + sqrt(1 - x)]

The product at numerator returns the difference of squares:

lim [(1  -x - h) - (1  -x)]/h*[sqrt (1 - x - h) + sqrt(1 - x)]

We'll eliminate like terms form numerator:

lim -h/h*[sqrt (1 - x - h) + sqrt(1 - x)]

We'll simplify and we'll get:

lim -1/[sqrt (1 - x - h) + sqrt(1 - x)]

We'll replace h by 0:

lim -1/[sqrt (1 - x - h) + sqrt(1 - x)] = -1/[sqrt (1 - x) + sqrt(1 - x)]

We'll combine like terms from denominator:

lim -1/[sqrt (1 - x - h) + sqrt(1 - x)] = -1/2sqrt (1 - x)

The first derivative of the given function, using the first principle, is f'(x) = -1/2sqrt (1 - x).