# FIND DERIVATIVES OF RADICAL FUNCTIONS

To find the derivative of a radical function, first write the radical sign as exponent and find derivative using chain rule.

Example :

f(x) = √x

f(x) = x1/2

Find derivative with respect to x.

f'(x) = (1/2)x1/2-1(x')

f'(x) = (1/2)x-1/2(1)

= (1/2)(1/x1/2)

= (1/2)(1/x)

= 1/(2x)

Base on the above example, we can derive formula for derivative of a radical function.

Let f(x) = √z. Then derivative f(x) :

Find the derivative of the following radical functions with respect to x :

Example 1 :

y = √(x + 2)

Solution :

y = √(x + 2)

y' = {1/[2√(x + 2)]}(x + 2)'

y' = {1/[2√(x + 2)]}(1)

y' = 1/[2√(x + 2)]

Example 2 :

y = √(2x - 1)

Solution :

y = √(2x - 2)

y' = {1/[2√(2x - 1)]}(2x - 1)'

y' = {1/[2√(2x - 1)]}(2)

y' = 1/√(2x - 1)

Example 3 :

y = √(3x2 + 5)

Solution :

y = √(3x2 + 5)

y' = {1/[2√(3x2 + 5)]}(3x2 + 5)'

y' = {1/[2√(3x2 + 5)]}(6x)

= 3x/√(3x2 + 5)

Example 4 :

y = √(2x4 + 2x - 1)

Solution :

y = √(2x4 + 2x - 1)

y' = {1/[2√(2x4 + 2x - 1)]}(2x4 + 2x - 1)'

y' = {1/[2√(2x4 + 2x - 1)]}(8x3 + 2)

= (4x3 + 1)/(2x4 + 2x - 1)

Example 5 :

y = (x3 + 2x)√x

Solution :

y = (x+ 2x)√x

Since two x terms are multiplied, we have to use the product rule to find the derivative.

 Let u = x3 + 2x.u' = 3x2 + 2(1)  = 3x2 + 2 Let v = √x.v' = 1/2√x

Product rule :

(uv)' = uv' + u'v

y' = (x+ 2x)(1/2√x) + (3x2 + 2)√x

= (x3/2√x + 2x/2√x) + 3x2√x + 2√x

= (1/2)x(3-1/2) + x(1 - 1/2) + 3x(2 + 1/2) + 2√x

= (1/2)x5/2 + x1/2 + 3x5/2 + 2√x

= [(1/2) + 3]x5/2 + √x + 2√x

= (7/2)x5/2 + 3√x

Example 6 :

y = (√x + 2x)/x2 - 1

Solution :

y = (√x + 2x)/x2 - 1

In the above function, we have variable x in both numerator and denominator.

So, we have to use the quotient rule to find the derivative

Quotient rule :

(u/v)' = (vu' - uv')/v2

 Let u = √x + 2x.u' = 1/2√x + 2(1)= 1/2√x + 2 Let v = x2 - 1.v' = 2x - 0= 2x

= [(x2 - 1)(1/2√x + 2) - (√x + 2x) (2x)]/(x2 - 1)2

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