Find derivatives of radical functions :
Here we are going to see how to find the derivatives of radical functions.
We use the formula given below to find the first derivative of radical function.
f(x) = √x
f'(x) = 1/(2√x)
Let us look into some example problems to understand the above concept.
Example 1 :
Find the derivative of the following function
y = (x^{3 }+ 2x) √x
Solution :
y = (x^{3 }+ 2x) √x
Since two x terms are multiplying, we have to use the product rule to find the derivative.
u = x^{3 }+ 2x u' = 3x^{2} + 2(1) = 3x^{2} + 2 |
v = √x v' = 1/2√x |
Product rule :
d (uv) = uv' + vu'
= (x^{3 }+ 2x)1/2√x + √x(3x^{2} + 2)
= (x^{3}/2√x^{ }+ 2x/2√x) + 3x^{2}√x + 2 √x
= (1/2)x^{(3-1/2)}^{ }+ x^{(1 - 1/2)} + 3x^{(}^{2 + 1/2)} + 2 √x
= (1/2)x^{5/2}^{ }+ x^{1/2} + 3x^{5/2} + 2 √x
= [(1/2) + 3]x^{5/2}^{ }+ √x + 2 √x
= (7/2) x^{5/2}^{ }+ 3 √x
Example 2 :
Find the derivative of the following function
y = (√x + 2x)/x^{2} - 1
Solution :
y = (√x + 2x)/x^{2} - 1
In the above question, In both numerator and denominator we have x functions.
So, we have to use the quotient rule to find the derivative
Quotient rule :
d (u/v) = (v u' - uv')/ v^{2}
u = √x + 2x
u' = (1/2 √x) + 2(1) ==> (1/2√x) + 2
v = x^{2} - 1
v' = 2x - 0 ==> 2x
= [(x^{2} - 1) ((1/2√x) + 2)) - (√x + 2x) (2x)] / (x^{2} - 1)^{2}
Example 3 :
Find the derivative of the following function
y = √(2x^{4} + 2x - 1)
Solution :
Let t = 2x^{4} + 2x - 1
y = √t
dy/dx = (dy/dt) (dt/dx)
dy/dt = 1/2√t dy/dx = 8x^{3} + 2(1) - 0
= 8x^{3} + 2
dy/dx = (1/2√t) (8x^{3} + 2)
By applying the value of t, we get
dy/dx = (1/2√2x^{4} + 2x - 1) (8x^{3} + 2)
= (4x^{3} + 1)/(√2x^{4} + 2x - 1)
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