Y=(4x+3)^4 (x+1)^-3 find the first and second derivatives&nbsp;

Question

Y=(4x+3)^4 (x+1)^-3 find the first and second derivatives&amp;nbsp;

star233 · Accepted Answer

To find the first and second derivatives of the given function, which is

Y = (4 x + 3)^{4} \cdot (x + 1)^{- 3}

, we will use the product rule and chain rule.
Let's start with the first derivative.
First Derivative:
To find the first derivative

Y^{'}

, we will apply the product rule:

Y = (4 x + 3)^{4} \cdot (x + 1)^{- 3}

Using the product rule, the first derivative is given by:

Y^{'} = {((4 x + 3)^{4})}^{'} \cdot (x + 1)^{- 3} + (4 x + 3)^{4} \cdot {((x + 1)^{- 3})}^{'}

Now, let's find the derivative of each term using the chain rule.
Derivative of

(4 x + 3)^{4}

:
We can rewrite this term as

u^{4}

, where

u = 4 x + 3

. Applying the chain rule, the derivative is:

{((4 x + 3)^{4})}^{'} = 4 \cdot (4 x + 3)^{3} \cdot (4 x + 3)^{'} = 4 \cdot (4 x + 3)^{3} \cdot 4 = 16 (4 x + 3)^{3}

Derivative of

(x + 1)^{- 3}

:
We can rewrite this term as

v^{- 3}

, where

v = x + 1

. Applying the chain rule, the derivative is:

{((x + 1)^{- 3})}^{'} = - 3 \cdot (x + 1)^{- 4} \cdot (x + 1)^{'} = - 3 \cdot (x + 1)^{- 4} \cdot 1 = - 3 (x + 1)^{- 4}

Substituting these derivatives back into the expression for

Y^{'}

, we have:

Y^{'} = 16 (4 x + 3)^{3} \cdot (x + 1)^{- 3} - 3 (4 x + 3)^{4} \cdot (x + 1)^{- 4}

Therefore, the first derivative of

Y

is:

Y^{'} = 16 (4 x + 3)^{3} \cdot (x + 1)^{- 3} - 3 (4 x + 3)^{4} \cdot (x + 1)^{- 4}

Second Derivative:
To find the second derivative

Y^{″}

, we will differentiate the first derivative

Y^{'}

with respect to

x

. Let's proceed:

Y^{'} = 16 (4 x + 3)^{3} \cdot (x + 1)^{- 3} - 3 (4 x + 3)^{4} \cdot (x + 1)^{- 4}

Applying the product rule and the chain rule, we can find the second derivative as follows:

Y^{″} = {(16 (4 x + 3)^{3} \cdot (x + 1)^{- 3})}^{'} - {(3 (4 x + 3)^{4} \cdot (x + 1)^{- 4})}^{'}

Differentiating each term using the chain rule, we obtain:
Derivative of

16 (4 x + 3)^{3}

:
Using the chain rule, the derivative is:

{(16 (4 x + 3)^{3})}^{'} = 3 \cdot 16 (4 x + 3)^{2} \cdot (4 x + 3)^{'} = 3 \cdot 16 (4 x + 3)^{2} \cdot 4 = 192 (4 x + 3)^{2}

Derivative of

(x + 1)^{- 3}

:
Using the chain rule, the derivative is:

{((x + 1)^{- 3})}^{'} = - 3 \cdot (x + 1)^{- 4} \cdot (x + 1)^{'} = - 3 \cdot (x + 1)^{- 4} \cdot 1 = - 3 (x + 1)^{- 4}

Derivative of

3 (4 x + 3)^{4}

:
Using the chain rule, the derivative is:

{(3 (4 x + 3)^{4})}^{'} = 4 \cdot 3 (4 x + 3)^{3} \cdot (4 x + 3)^{'} = 12 (4 x + 3)^{3} \cdot 4 = 48 (4 x + 3)^{3}

Derivative of

(x + 1)^{- 4}

:
Using the chain rule, the derivative is:

{((x + 1)^{- 4})}^{'} = - 4 \cdot (x + 1)^{- 5} \cdot (x + 1)^{'} = - 4 \cdot (x + 1)^{- 5} \cdot 1 = - 4 (x + 1)^{- 5}

Substituting these derivatives back into the expression for

Y^{″}

, we have:

Y^{″} = 192 (4 x + 3)^{2} \cdot (x + 1)^{- 3} + 16 (4 x + 3)^{3} \cdot (x + 1)^{- 4} - 48 (4 x + 3)^{3} \cdot (x + 1)^{- 4} - 12 (4 x + 3)^{4} \cdot (x + 1)^{- 5}

Simplifying further, we can combine like terms:

Y^{″} = 192 (4 x + 3)^{2} \cdot (x + 1)^{- 3} - 32 (4 x + 3)^{3} \cdot (x + 1)^{- 4} - 12 (4 x + 3)^{4} \cdot (x + 1)^{- 5}

Therefore, the second derivative of

Y

is:

Y^{″} = 192 (4 x + 3)^{2} \cdot (x + 1)^{- 3} - 32 (4 x + 3)^{3} \cdot (x + 1)^{- 4} - 12 (4 x + 3)^{4} \cdot (x + 1)^{- 5}

Y=(4x+3)^4 (x+1)^-3 find the first and second derivatives