# What is the first derivative of function given by f(x)=(x-square rootx)/(5x^2-3) ?

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### 2 Answers

We have to find the derivative of (x - sqrt x) / (5x^2 - 3).

As we have the expression as a quotient, we use the quotient rule.

f(x) = u(x) / v(x)

=> f'(x) = (u'(x)*v(x) - u(x)*v'(x))/v(x)^2

Here, u(x) = x - sqrt x, v(x) = 5x^2 - 3

u'(x) = 1 - (1/2)(1/sqrt x) = 1 - 1/(2*sqrt x) = (2*sqrt x - 1)/(2*sqrt x)

v'(x) = 10x

f'(x) = [(5x^2 - 3)(2*sqrt x - 1)/(2*sqrt x) - (x - sqrt x)*10x]/(5x^2 - 3)^2

f'(x) = [(5x^2 - 3)(2*sqrt x - 1) - 10x*(x - sqrt x)*(2*sqrt x)] / (2*sqrt x)*(5x^2 - 3)^2

f'(x) = [sqrt x*(10x^2 - 6) - 5x^2 + 3 - 20x^2*sqrt x + 20x^2] / (2*sqrt x)*(5x^2 - 3)^2

f'(x) = [-sqrt x*(10x^2 + 6) + 15x^2 + 3] / (2*sqrt x)*(5x^2 - 3)^2

f'(x) = - [sqrt x*(10x^2 + 6) - 15x^2 - 3] / (2*sqrt x)*(5x^2 - 3)^2

**The required derivative is [-sqrt x*(10x^2 + 6) + 15x^2 + 3] / [(2*sqrt x)*(5x^2 - 3)^2]**

Since the given function is a quotient of 2 functions, we'll apply the quotient rule:

(u/v)' = (u'*v - u*v')/v^2

We'll put u = x - sqrtx

We'll differentiate with respect to x:

u' = 1 - 1/2sqrtx

We'll put v = 5x^2-3

We'll differentiate with respect to x:

v' = 10x

We'll apply quotient rule:

f'(x) = [(1 - 1/2sqrtx)(5x^2-3)-(x - sqrtx)(10x)]/(5x^2-3)^2

f'(x) = [(2sqrtx - 1)(5x^2-3) - 10x^2*2sqrtx + 20x^2]/(5x^2-3)^2

**f'(x) = [(2sqrtx - 1)(5x^2-3) - 20x*sqrtx*(x - sqrtx)]/(5x^2-3)^2**