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How do you differentiate #x^(2/3)+y^(2/3)=pi^(2/3)#?

Answer 1

Differential is #-root(3)(x/y)#

Differential of #x^n# is #nx^(n-1)# and that of #y^n#, is #ny^(n-1)(dy)/(dx)# using implicit differentiation. Also as #pi# is a constant, its differential, or that of any of its power, is #0#

Hence differential of #x^(2/3)+y^(2/3)=pi^(2/3)#

is #2/3x^(-1/3)+2/3y^(-1/3)(dy)/(dx)=0#

and #(dy)/(dx)=-(2/3y^(-1/3))/(2/3x^(-1/3))=-(1/root(3)y)/(1/root(3)x)#

= #-root(3)(x/y)#

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Answer 2

Ezra Adams

To differentiate the equation (x^{2/3} + y^{2/3} = \pi^{2/3}) with respect to (x), you can follow these steps:

Rewrite the equation as (y = (\pi^{2/3} - x^{2/3})^{3/2}).
Differentiate (y) with respect to (x) using the chain rule and power rule.
The result is (\frac{dy}{dx} = -\frac{1}{\sqrt{2}}\left(\pi^{2/3} - x^{2/3}\right)^{1/2} \cdot \frac{2}{3}x^{-1/3}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.