Question

find the nth derivative of sqrt (ax+b)​

Answer 1

Step-by-step explanation:

Given that

$y=\sqrt\left ( ax+b \right )$

Lets find the first derivative

$\dfrac{dy}{dx}=\dfrac{1}{2}a\left ( ax+b \right )^{-\dfrac{1}{2}}$

Now find second derivative

$\dfrac{d^2y}{dx^2}=-\dfrac{1}{4}a^2\left ( ax+b \right )^{-\dfrac{3}{2}}$

$\dfrac{d^3y}{dx^3}=\dfrac{3}{8}a^3\left ( ax+b \right )^{-\dfrac{5}{2}}$

$\dfrac{d^4y}{dx^4}=-\dfrac{15}{16}a^4\left ( ax+b \right )^{-\dfrac{7}{2}}$

And so on.

Answer 2

Step-by-step explanation:

Given that

y=\sqrt ( ax+b )y=

(

ax+b)

Lets find the first derivative

\dfrac{dy}{dx}=\dfrac{1}{2}a ( ax+b )^{-\dfrac{1}{2}}

dx

dy

=

2

1

a(ax+b)

−

2

1

Now find second derivative

\dfrac{d^2y}{dx^2}=-\dfrac{1}{4}a^2 ( ax+b )^{-\dfrac{3}{2}}

dx

2

d

2

y

=−

4

1

a

2

(ax+b)

−

2

3

\dfrac{d^3y}{dx^3}=\dfrac{3}{8}a^3 ( ax+b )^{-\dfrac{5}{2}}

dx

3

d

3

y

=

8

3

a

3

(ax+b)

−

2

5

\dfrac{d^4y}{dx^4}=-\dfrac{15}{16}a^4 ( ax+b )^{-\dfrac{7}{2}}

dx

4

d

4

y

=−

16

15

a

4

(ax+b)

−

2

7

And so on.