Calculate partial derivative from first principles: f(x,y) = (x^3+y^3)^1/3

[evantheman]

Hey guys,

I'm having trouble with this question, using this f(x,y) = (x^3+y^3)^1/3, it's asking to calculate the first partial derivative with respect to x from first principles. Could somebody shine some light upon this question. I understand subbing in x+h etc but don't get how I can cancel out parts of the equation to get the actual derivative with respect to x.

Thanks

 

[skaa]

\(\displaystyle \frac{\partial f}{\partial x}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{((x+h)^3+y^3)^{\frac{1}{3}}-(x^3+y^3)^{\frac{1}{3}}}{h}=\)

\(\displaystyle =\lim_{h\to 0}\frac{(((x+h)^3+y^3)^{\frac{1}{3}}-(x^3+y^3)^{\frac{1}{3}})(((x+h)^3+y^3)^{\frac{2}{3}}+((x+h)^3+y^3)^{\frac{1}{3}}(x^3+y^3)^{\frac{1}{3}}+(x^3+y^3)^{\frac{2}{3}})}{h(((x+h)^3+y^3)^{2/3}+((x+h)^3+y^3)^{\frac{1}{3}}(x^3+y^3)^{\frac{1}{3}}+(x^3+y^3)^{\frac{2}{3}})}=(*)\)

Now we will use this:
\(\displaystyle (a-b)(a^2+ab+b^2)=(a-b)^3\)

\(\displaystyle (*)=\lim_{h\to 0}\frac{((x+h)^3+y^3)-(x^3+y^3)}{h(((x+h)^3+y^3)^{\frac{2}{3}}+((x+h)^3+y^3)^{\frac{1}{3}}(x^3+y^3)^{\frac{1}{3}}+(x^3+y^3)^{\frac{2}{3}})}=\)

\(\displaystyle =\lim_{h\to 0}\frac{3x^2h+3xh^2+h^3}{h((x^3+y^3)^{\frac{2}{3}}+(x^3+y^3)^{\frac{1}{3}}(x^3+y^3)^{\frac{1}{3}}+(x^3+y^3)^{\frac{2}{3}})}=\frac{3x^2}{3(x^3+y^3)^{2/3}}=\frac{x^2}{(x^3+y^3)^{\frac{2}{3}}}\)

 

[evantheman]

\(\displaystyle \frac{\partial f}{\partial x}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{((x+h)^3+y^3)^{\frac{1}{3}}-(x^3+y^3)^{\frac{1}{3}}}{h}=\)

\(\displaystyle =\lim_{h\to 0}\frac{(((x+h)^3+y^3)^{\frac{1}{3}}-(x^3+y^3)^{\frac{1}{3}})(((x+h)^3+y^3)^{\frac{2}{3}}+((x+h)^3+y^3)^{\frac{1}{3}}(x^3+y^3)^{\frac{1}{3}}+(x^3+y^3)^{\frac{2}{3}})}{h(((x+h)^3+y^3)^{2/3}+((x+h)^3+y^3)^{\frac{1}{3}}(x^3+y^3)^{\frac{1}{3}}+(x^3+y^3)^{\frac{2}{3}})}=(*)\)

Now we will use this:
\(\displaystyle (a-b)(a^2+ab+b^2)=(a-b)^3\)

\(\displaystyle (*)=\lim_{h\to 0}\frac{((x+h)^3+y^3)-(x^3+y^3)}{h(((x+h)^3+y^3)^{\frac{2}{3}}+((x+h)^3+y^3)^{\frac{1}{3}}(x^3+y^3)^{\frac{1}{3}}+(x^3+y^3)^{\frac{2}{3}})}=\)

\(\displaystyle =\lim_{h\to 0}\frac{3x^2h+3xh^2+h^3}{h((x^3+y^3)^{\frac{2}{3}}+(x^3+y^3)^{\frac{1}{3}}(x^3+y^3)^{\frac{1}{3}}+(x^3+y^3)^{\frac{2}{3}})}=\frac{3x^2}{3(x^3+y^3)^{2/3}}=\frac{x^2}{(x^3+y^3)^{\frac{2}{3}}}\)

Mate, thank you so you much!!!

 

[CalicoCat]

At (*), Why do you times by (((x+h)^3+y^3)^(2/3)+((x+h)^3+y^3)^(1/3)(x3+y3)^(1/3)+(x3+y3)^(2/3)) on the top and bottom? Like why not just multiple everything to the power of 3 instead?

e.g. ((x^3+y^3)^(1/3))^3 --> x^3+y^3

I am slightly confused by your steps. Are you able to explain what you're doing more?

 

[HallsofIvy]

At (*), Why do you times by (((x+h)^3+y^3)^(2/3)+((x+h)^3+y^3)^(1/3)(x3+y3)^(1/3)+(x3+y3)^(2/3)) on the top and bottom? Like why not just multiple everything to the power of 3 instead?

e.g. ((x^3+y^3)^(1/3))^3 --> x^3+y^3

I am slightly confused by your steps. Are you able to explain what you're doing more?

\(\displaystyle (a^{1/3})^3= a\) but \(\displaystyle (a^{1/3}- b^{1/3})^3= a- 3a^{2/3}b^{1/3}+ 3a^{1/3}b^{2/3}- b\), NOT \(\displaystyle a- b\).


Instead skaa is using the fact that \(\displaystyle (a- b)(a^2+ ab+ b^2)= a^3- b^3\) with \(\displaystyle a= x^{1/3}\) and \(\displaystyle b= y^{1/3}\).

By the way, you multiply two numbers, you do not "times" them!