🔒 Closed Pa help po calculus

[Rojen G Embate]

Pwede po makahingi solutions nito para makapag aralan ko po

 

[Slime-]

To differentiate Y with respect to x, we can use the power rule and the chain rule of differentiation.

Y = x + 1/√x

Y' = d/dx(x) + d/dx(1/√x)

Applying the power rule and chain rule:

Y' = 1 + (-1/2)x^(-3/2)

Simplifying the second term:

Y' = 1 - 1/(2x^(3/2))

Therefore, the derivative of Y with respect to x is:

Y' = 1 - 1/(2x^(3/2))

To differentiate Y with respect to x, we can use the power rule and the chain rule of differentiation.

Y = x⁴ + √x

Y' = d/dx(x⁴) + d/dx(√x)

Applying the power rule and chain rule:

Y' = 4x³ + (1/2)x^(-1/2)

Simplifying the second term:

Y' = 4x³ + 1/(2√x)

Therefore, the derivative of Y with respect to x is:

Y' = 4x³ + 1/(2√x)

------
Can you get the 2nd and 3rd order differential of this: Y= x+1/√x

Sure, we can find the 2nd and 3rd order derivatives of Y with respect to x using the power rule and the chain rule of differentiation.

First, let's find the first derivative of Y:

Y = x + 1/√x

Y' = 1 - 1/(2x^(3/2))

Next, we can differentiate Y' to find the second derivative:

Y'' = d/dx(1 - 1/(2x^(3/2)))

Using the power rule and chain rule:

Y'' = 3/(4x^(5/2))

Therefore, the second derivative of Y with respect to x is:

Y'' = 3/(4x^(5/2))

To find the third derivative of Y, we differentiate Y'' with respect to x:

Y''' = d/dx(3/(4x^(5/2)))

Using the power rule and chain rule:

Y''' = -15/(8x^(7/2))

Therefore, the third derivative of Y with respect to x is:

Y''' = -15/(8x^(7/2))

 

[Rojen G Embate]

To differentiate Y with respect to x, we can use the power rule and the chain rule of differentiation.

Y = x + 1/√x

Y' = d/dx(x) + d/dx(1/√x)

Applying the power rule and chain rule:

Y' = 1 + (-1/2)x^(-3/2)

Simplifying the second term:

Y' = 1 - 1/(2x^(3/2))

Therefore, the derivative of Y with respect to x is:

Y' = 1 - 1/(2x^(3/2))

To differentiate Y with respect to x, we can use the power rule and the chain rule of differentiation.

Y = x⁴ + √x

Y' = d/dx(x⁴) + d/dx(√x)

Applying the power rule and chain rule:

Y' = 4x³ + (1/2)x^(-1/2)

Simplifying the second term:

Y' = 4x³ + 1/(2√x)

Therefore, the derivative of Y with respect to x is:

Y' = 4x³ + 1/(2√x)

------
Can you get the 2nd and 3rd order differential of this: Y= x+1/√x

Sure, we can find the 2nd and 3rd order derivatives of Y with respect to x using the power rule and the chain rule of differentiation.

First, let's find the first derivative of Y:

Y = x + 1/√x

Y' = 1 - 1/(2x^(3/2))

Next, we can differentiate Y' to find the second derivative:

Y'' = d/dx(1 - 1/(2x^(3/2)))

Using the power rule and chain rule:

Y'' = 3/(4x^(5/2))

Therefore, the second derivative of Y with respect to x is:

Y'' = 3/(4x^(5/2))

To find the third derivative of Y, we differentiate Y'' with respect to x:

Y''' = d/dx(3/(4x^(5/2)))

Using the power rule and chain rule:

Y''' = -15/(8x^(7/2))

Therefore, the third derivative of Y with respect to x is:

Y''' = -15/(8x^(7/2))

wala po bang pina easy method jan yong walang mga ganto po ^, /, )) di ko po ma gets

 

[Slime-]

wala po bang pina easy method jan yong walang mga ganto po ^, /, )) di ko po ma gets

Bigay lang din ni ᑕᕼᗩTGᑭT yan hahaha

Ito may explanation nya

You do not have permission to view the full content of this post. Log in or register now.

 

[Rojen G Embate]

Bigay lang din ni ᑕᕼᗩTGᑭT yan hahaha

Ito may explanation nya

You do not have permission to view the full content of this post. Log in or register now.

lods pano naman sa number 3 at 4