Differentiate y=x[twelve]/[thirteen] two ways using

[TONYYEUNG]

Please help me to answer the following question:

Differentiate y=x[twelve]/[thirteen] two ways using

(i) implicit differentiation; and

My answer:

y[thirteen]=x[twelve]

So, 13y[twelve]dy/dx=12x[eleven]

Could you tell me how can I do next to the solution?? And could you tell me how can I do part (ii)??

(ii) the Chain Rule

 

[skeeter]

sorry for being so ignorant, but what is y=x[twelve]/[thirteen] ?

do you mean \(\displaystyle \L x^{\frac{12}{13}}\) ???

 

[TONYYEUNG]

Yes, please help me.

Yes, please help me.

 

[soroban]

Hello, Tony!

\(\displaystyle \text{Differentiate }y\:=\:x^{\frac{12}{13}}\)

[1] by implicit differentiation and [2] by the Chain Rule.

It's a strange problem . . . I'll try to comply.


[1] Raise both sides to the 13^th power: \(\displaystyle \:y^{13}\:=\:x^{12}\)

Differentiate implicitly: \(\displaystyle \:13y^{12}\left(\frac{dy}{dx}\right)\:=\:12x^{11}\)

. . Therefore: \(\displaystyle \L\:\frac{dy}{dx} \:=\:\frac{12x^{11}}{13y^{12}}\)


[2] We have: \(\displaystyle \:y\:=\:\left(x^{12}\right)^{\frac{1}{13}}\)

Chain Rule: \(\displaystyle \:\frac{dy}{dx}\:=\:\frac{1}{13\,}\left(x^{12}\right)^{-\frac{12}{13}}\,\left(12x^{11}\right)\:=\:\frac{12}{13}\left(x^{-\frac{144}{13}}\right)\left(x^{11}\right)\)

. . Therefore: \(\displaystyle \L\:\frac{dy}{dx}\:=\:\frac{12}{13}x^{-\frac{1}{13}}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Part [2] can be done like this, too . . .

We have: \(\displaystyle \:y\:=\:\left(x^{\frac{1}{13}}\right)^{12}\)

Then: \(\displaystyle \:\frac{dy}{dx}\:=\:12\left(x^{\frac{1}{13}}\right)^{11}\left(\frac{1}{13}x^{-\frac{12}{13}}\right) \:=\:\frac{12}{13}\left(x^{\frac{11}{13}}\right)\left(x^{-\frac{12}{13}}\right)\)

Therefore: \(\displaystyle \L\:\frac{dy}{dx}\:=\:\frac{12}{13}x^{-\frac{1}{13}}\) . . . ta-DAA!