Implicit differentiation, whats going wrong?

[luckyducky87]

Implicit differentiation, what's going wrong!?

Hey people can someone point out to me please where I'm going wrong with part B of this question, can't get it to look the answer in the book, d2y/dx2 = -3(x^6/y^7)- 3(x^2/y^3).

My answer is listed below under part B section, but i can't manipulate it to look like the above answer :S, any help/tips greatly appreaciated cheers,

5. Implicit Differentiation. If x^4 + y^4 = 16, use the following steps to find y''.

(a) Use implicit differentiation to find y',

dy/dx = -x^(3)/y^(3) - too easy

(b) Use the quotient or product rule to differentiate the expression for y' from part (a). Express your answer in terms of x and y only.

d^2/dx^2= ((-x^(3))/(-3y^(4)))*y''-(3x^(2))/(y^(3))

(c) Use the fact that x and y must satisfy x4 +y4 = 16 to simplify your answer to part

(b) to the following expression

d2y/dx2 = -48(x2/y7).

 

[luckyducky87]

The answer is d2y/dx2= -(3x^6/y^7) -(3x^2/y^3), cannot manipulate dy/dx = -x^3/y^3 to look like it though, any help or tips appreciated thanks,

lucky.

 

[HallsofIvy]

Hey people can someone point out to me please where I'm going wrong with part B of this question, can't get it to look the answer in the book, d2y/dx2 = -3(x^6/y^7)- 3(x^2/y^3).

My answer is listed below under part B section, but i can't manipulate it to look like the above answer :S, any help/tips greatly appreaciated cheers,

5. Implicit Differentiation. If x^4 + y^4 = 16, use the following steps to find y''.

(a) Use implicit differentiation to find y',

dy/dx = -x^(3)/y^(3) - too easy

Okay

(b) Use the quotient or product rule to differentiate the expression for y' from part (a). Express your answer in terms of x and y only.

d^2/dx^2= ((-x^(3))/(-3y^(4)))*y''-(3x^(2))/(y^(3))

I have no idea how you got that. In particular, I don't see how you got a y'' on the right when there was no y' on the right to begin with.

[tex]\frac{d^2y}{dx^2}= \frac{-3x^2y^3- (-x^3)(3y^2)y'}{y^6}[/itex]
by the quotient rule. Now, to "express your answer in terms of x and y only", replace that y' with $-x^3/y^3$:
$$\frac{d^2y}{dx^2}= \frac{-3x^2y^3+ 3x^3y^2(-x^3/y^3)}{y^6}$$
[tex]= \frac{-3x^2y^3- 3x^6y^{-1}}{y^6}= -3x^2\frac{y^4+ x^4}{y^7}[/itex]

(c) Use the fact that x and y must satisfy x4 +y4 = 16 to simplify your answer to part

(b) to the following expression

d2y/dx2 = -48(x2/y7).

$$-3x^2\frac{y^4+ x^4}{y^7}$$
becomes
$$-3x^2\frac{16}{y^7}$$

 

[luckyducky87]

Thanks, appreciate the help,

Lucky