DERIVATIVES OF RADICAL FUNCTIONS WORKSHEET

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

1. y = √(x + 2)

2. y = √(2x - 1)

3. y = √(3x² + 5)

4. y = √(2x⁴ + 2x - 1)

5. y = (x³ + 2x)√x

6. y = (√x + 2x)/x² - 1

7. ⁴√(7m³ - 4m² + 2)

8. y = (x - 3)√(x - 3)

9. y = √(4 - √(x - 2))

10.  log (2x² - 1)/√x

11. Find the equation of tangent to the curve at the point (1, e).

y = ln √(2 - x)

12. y = √cos 2x

13. y = tan √cos x

1. Answer :

y = √(x + 2)

y' = {1/[2√(x + 2)]}(x + 2)'

y' = {1/[2√(x + 2)]}(1)

y' = 1/[2√(x + 2)]

2. Answer :

y = √(2x - 2)

y' = {1/[2√(2x - 1)]}(2x - 1)'

y' = {1/[2√(2x - 1)]}(2)

y' = 1/√(2x - 1)

3. Answer :

y = √(3x² + 5)

y' = {1/[2√(3x² + 5)]}(3x² + 5)'

y' = {1/[2√(3x² + 5)]}(6x)

= 3x/√(3x² + 5)

4. Answer :

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)

5. Answer :

y = (x³ + 2x)√x

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

Product rule :

(uv)' = uv' + u'v

y' = (x³ + 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)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

6. Answer :

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

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

7. Answer :

 ⁴√(7m³ - 4m² + 2)

Since we have 4^th root here, we write it down in exponential form and then using chain rule we find the derivative.

y = (7m³ - 4m² + 2)^1/4

= (1/4) (7m³ - 4m² + 2)^{(1/4) - 1} [7(3m²) - 4(2m) + 0]

= (1/4) (7m³ - 4m² + 2)^{(1 - 4)/4} [21m² - 8m]

= (1/4) (7m³ - 4m² + 2)^-3/4 [21m² - 8m]

8. Answer :

y = (x - 3)√(x - 3)

Since these two functions are multiplied, we have to use product rule to find the derivative.

u = x - 3 and v = √(x - 3)

u' = 1 - 0 and v' = 1/2√(x - 3)

u' = 1 and v' = 1/2√(x - 3)

d(uv) = uv' + vu'

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

= [(x - 3) + 2√(x - 3)√(x - 3)] / 2√(x - 3)

= [(x - 3) + 2(x - 3)] / 2√(x - 3)

= (x - 3 + 2x - 6) / 2√(x - 3)

= (3x - 9) / 2√(x - 3)

= 3(x - 3) / 2√(x - 3)

9. Answer :

y = √(4 - √(x - 2))

Squring on both sides

y² = 4 - √(x - 2)

Differentiating with respect to x, we get

2y(dy/dx) = 0 - 1/2√(x - 2) (1)

2y(dy/dx) = - 1/2√(x - 2)

dy/dx = (1/2y) (-1/2√(x - 2))

= -1/4y√(x - 2)

Applying the value of y, we get

= -1/4√(x - 2) √(4 - √(x - 2)) 

10. Answer :

Let y = log (2x² - 1)/√x

Using the rules of logarithms,

log (m/n) = log m - log n

y = log (2x² - 1) - log √x

Differentitating with respect to x, we get

dy/dx = 1/(2x² - 1) (4x - 0) - 1/2√x

= 4x/(2x² - 1) - 1/2√x

11. Answer :

Find the equation of tangent to the curve at all points.

y = ln √(2 - x)

dy/dx = 1/2√(2 - x) (0 - 1)

Slope = -1/2√(2 - x)

Equation of tangent at (1, e)

Slope at (1, e) = -1/2√(2 - 1)

= -1/2√1

= -1/2

Equation of tangent :

(y - y₁) = m(x - x₁)

y - e = -1/2 (x - 1)

y = -x/2 + 1/2 + e

12. Answer :

y = √cos 2x

Differentiate with respect to x

dy/dx = (1/2√cos 2x) (-sin 2x) (2)

dy/dx = (-2(sin 2x)/2√cos 2x) 

dy/dx = (-sin 2x)/√cos 2x

= (-sin 2x)/√cos 2x

13. Answer :

y = tan √cos x

Differentiate with respect to x,

dy/dx = sec² (√cos x) (1/2√cos x) (-sin x)

= (-sin x)sec² (√cos x)/2√cos x

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