Implicit differentiation

RM5152

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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 :)
 

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Up until this point you have y4y^4 in the denominator.
Screen Shot 2022-06-29 at 10.56.46 AM.png

After subbing yy' it becomes y5?y^5?
Screen Shot 2022-06-29 at 10.57.47 AM.png
 
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.
2xy22x4y1y4Multiply top and bottom by y:2xy22x4y1y4yyYou’re missing distributive property.2xy2y2x4y1yy4y2xy32x4y5\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}\\So it should be y3y^3.
 
Why do I multiply the top and bottom by y? Thanks for the help
Multiply by y gets rid of negative exponents y1y^{-1} in the numerator. Isn't that what you're trying to do?
2xy22x4y1y4\frac{-2xy^2-2x^4y^{-1}}{y^4}\\ This would've been acceptable as well. Just a matter of preference.
 
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
 
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.
62215\displaystyle \frac{6-2\cdot 2^{-1}}{5}
Say you want to get rid of the 212^{-1}.
How would you do it? Check your answer with a calculator if your method is correct.
The correct answer we know is 1.
 
Take a more concrete example.
62215\displaystyle \frac{6-2\cdot 2^{-1}}{5}
Say you want to get rid of the 212^{-1}.
How would you do it? Check your answer with a calculator if your method is correct.
The correct answer we know is 1.
x2y2+x4y\displaystyle x^2 * y^2 + \frac{x^4}{y}

= [x2y2]yy+x4y\displaystyle [x^2 * y^2] * \frac{y}{y} + \frac{x^4}{y}

= [x2y3]+x4y\displaystyle \frac {[x^2 * y^3] + x^4}{y}
 
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