Implicit differentiation

[RM5152]

Hello.. I’m asked to find the second derivative (y’’(x)) at the point (3,1) when x^3+y^3=28. I have done the workings up until the point where I know I should be isolating (x^3+y^3) and substituting in 28 but I can’t see how the y component becomes y^3 .. I keep getting -2x(y^2-x^3)/y^5 ….. I saw on another forum that other people are getting -2x(y^3+x^3) … where am I going wrong in terms of having y^2 instead of y^3 ?? Thanks for any help .. also if anyone could send me in the direction of a good video for learning implicit differentiation thay would be amazing, my teacher only went over it very briefly so I need to learn it properly outside of class. Thanks so much

 

[BigBeachBanana]

Up until this point you have [imath]y^4[/imath] in the denominator.


After subbing [imath]y'[/imath] it becomes [imath]y^5?[/imath]

 

[RM5152]

Up until this point you have [imath]y^4[/imath] in the denominator.
View attachment 33245

After subbing [imath]y'[/imath] it becomes [imath]y^5?[/imath]
View attachment 33246

y’=-x^2/y^2 so I multiply the top line by (-x^2)(y^-2) and this y^-2 multiplied by y becomes y^-1 and then I bring this down to the bottom line which makes y^4*y=y^5 … let me know if this is wrong. Sorry that it’s confusing

 

[BigBeachBanana]

y’=-x^2/y^2 so I multiply the top line by (-x^2)(y^-2) and this y^-2 multiplied by y becomes y^-1 and then I bring this down to the bottom line which makes y^4*y=y^5 … let me know if this is wrong. Sorry that it’s confusing

You made an algebraic mistake.
[math]\frac{-2xy^2-2x^4y^{-1}}{y^4}\\ \text{Multiply top and bottom by y:}\\ \frac{-2xy^2-2x^4y^{-1}}{y^4}\cdot \frac{\red{y}}{\red{y}}\\ \text{You're missing distributive property.}\\ \frac{-2xy^2\red{y}-2x^4y^{-1}\red{y}}{y^4\red{y}}\\ \frac{-2xy^3-2x^4}{y^5}\\[/math]So it should be [imath]y^3[/imath].

 

[RM5152]

Why do I multiply the top and bottom by y? Thanks for the help

 

[BigBeachBanana]

Why do I multiply the top and bottom by y? Thanks for the help

Multiply by y gets rid of negative exponents [imath]y^{-1}[/imath] in the numerator. Isn't that what you're trying to do?
[imath]\frac{-2xy^2-2x^4y^{-1}}{y^4}\\[/imath] This would've been acceptable as well. Just a matter of preference.

 

[RM5152]

I thought the y^-1 on the numerator could just become 1/y so the y just goes to the denominator.. so do I have to always multiply the top and bottom by y when there is a negative power on the top? Thanks

 

[BigBeachBanana]

I thought the y^-1 on the numerator could just become 1/y so the y just goes to the denominator.. so do I have to always multiply the top and bottom by y when there is a negative power on the top? Thanks

Take a more concrete example.
\(\displaystyle \frac{6-2\cdot 2^{-1}}{5}\)
Say you want to get rid of the [imath]2^{-1}[/imath].
How would you do it? Check your answer with a calculator if your method is correct.
The correct answer we know is 1.

 

[Deleted member 4993]

Take a more concrete example.
\(\displaystyle \frac{6-2\cdot 2^{-1}}{5}\)
Say you want to get rid of the [imath]2^{-1}[/imath].
How would you do it? Check your answer with a calculator if your method is correct.
The correct answer we know is 1.

\(\displaystyle x^2 * y^2 + \frac{x^4}{y}\)

= \(\displaystyle [x^2 * y^2] * \frac{y}{y} + \frac{x^4}{y}\)

= \(\displaystyle \frac {[x^2 * y^3] + x^4}{y}\)

 

[Steven G]

8 + 3*5^-1 = (8+3)/5 ??????

 

[Deleted member 4993]

8 + 3*5^-1 = (8+3)/5 ??????

To which post are you respond ing ?