Find d^2y/dx^2 of 3x=y^2-x^2

Calculus
Tutor: None Selected Time limit: 1 Day

Given 3x = y^2 - x^2 , find d^2y/dx^2

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Oct 3rd, 2015

3x = y^2 - x^2

Now

y^2 = x^2 + 3x

Taking derivative w.r.t x, then

d/dx(y^2) = d/dx(x^2 + 3x)

2y dy/dx = 2x + 3

Now

dy/dx = (2x + 3)/2y

As

y = √ (x^2 + 3x) so

dy/dx = (2x + 3)/(√(x^2 + 3x))

Again taking derivative w.r.t x, we get

d/dx(dy/dx) = d/dx[ (2x + 3) / √(x^2 + 3x) ]

Using the rule of d/dx(u/v) = [v*d/dx(u) - u*d/dx(v)] / v^2

d^2 y/dx^2 = [ √(x^2 + 3x) * d/dx(2x + 3) - (2x + 3) * d/dx(√(x^2 + 3x)) ] / (√x^2 + 3x)^2

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

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

                   = [ {4(x^2 + 3x) - (2x + 3)^2} / 2√(x^2 + 3x) ] / (x^2 + 3x)

                  = (4x^2 + 12x - 4x^2 - 9 - 12x) / (x^2 + 3x) * 2√(x^2 + 3x)

                  = - 9 / 2(x^2 + 3x)^3/2

Hence         d^2 y/dx^2 = - 9 / 2(x^2 + 3x)^3/2

Sep 22nd, 2015

The answer should be -9/4y^3 

It says in my textbook

Sep 22nd, 2015

The answer should be -9/4y^3 

It says in my textbook

Sep 22nd, 2015

okay the answer we got is - 9 / 4(x^2 + 3x)^3/2

And we also know that 

x^2 + 3x = y^2

So put the value of x^2 + 3x in the answer, we get

- 9 / 4(y^2)^3/2

- 9 / 4(y^2*^3/2)

- 9 / 4(y^3)

- 9 / 4y^3

Hence 

Sep 22nd, 2015

Oh okay, thanks a bunch!

Sep 22nd, 2015

3x = y^2 - x^2

Now

y^2 = x^2 + 3x

Taking derivative w.r.t x, then

d/dx(y^2) = d/dx(x^2 + 3x)

2y dy/dx = 2x + 3

Now

dy/dx = (2x + 3)/2y

As

y = √ (x^2 + 3x) so

dy/dx = (2x + 3)/(2√(x^2 + 3x))

Again taking derivative w.r.t x, we get

d/dx(dy/dx) = d/dx[ (2x + 3) / 2√(x^2 + 3x) ]

Using the rule of d/dx(u/v) = [v*d/dx(u) - u*d/dx(v)] / v^2

d^2 y/dx^2 = [ 2√(x^2 + 3x) * d/dx(2x + 3) - (2x + 3) * d/dx(2√(x^2 + 3x)) ] / (2√x^2 + 3x)^2

                   = [ 2*(2√x^2 + 3x) - (2x + 3)(1/√(x^2 + 3x) * (2x + 3)) ] / 4(x^2 + 3x)

                   = [ 4√(x^2 + 3x) - (2x + 3)^2 / √(x^2 + 3x) ] / 4(x^2 + 3x)

                   = [ {4(x^2 + 3x) - (2x + 3)^2} / √(x^2 + 3x) ] / 4(x^2 + 3x)

                  = (4x^2 + 12x - 4x^2 - 9 - 12x) / (x^2 + 3x) * 4√(x^2 + 3x)

                  = - 9 / 4(x^2 + 3x)^3/2

Hence         d^2 y/dx^2 = - 9 / 4(x^2 + 3x)^3/2

And we also know that 

x^2 + 3x = y^2

So put the value of x^2 + 3x in the answer, we get

- 9 / 4(y^2)^3/2

- 9 / 4(y^2*^3/2)

- 9 / 4(y^3)

- 9 / 4y^3

Hence 

Sep 22nd, 2015

There was a mistake and I corrected it in above. Okay???

Sep 22nd, 2015

Thats fine, thank you so much!

Sep 22nd, 2015

My Pleasure!!!

Sep 22nd, 2015

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