How do you find #(dy)/(dx)# given #root3x+root3(y^4)=2#?

Answer 1

Start by writing the roots as rational exponents:

#x^(1/3) + y^(4/3) = 2#

Differentiate each term using the power rule:

#1/3x^(1/3 - 1) + 4/3y^(4/3 - 1)dy/dx = d/dx(2)#
#1/3x^(-2/3) + 4/3y^(1/3) dy/dx = 0#
#1/(3x^(2/3)) + 4/3y^(1/3)dy/dx = 0#
#4/3y^(1/3)dy/dx = 0- 1/(3x^(2/3))#
#dy/dx = (- 1/(3x^(2/3)))/(4/3y^(1/3))#
#dy/dx = ( - 1)/(3x^(2/3)) xx 1/(4/3y^(1/3))#
#dy/dx = - 1/(4x^(2/3)y^(1/3))#

If you want it in radical form instead of rational exponent form:

#dy/dx = - 1/(4root(3)(x^2)root(3)(y))#

Hopefully this helps!

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Answer 2

To find (\frac{{dy}}{{dx}}), we differentiate the equation (\sqrt{3x} + \sqrt{3y^4} = 2) implicitly with respect to (x) using the chain rule and product rule. After differentiation, we solve for (\frac{{dy}}{{dx}}).

Differentiating (\sqrt{3x} + \sqrt{3y^4} = 2) with respect to (x) yields:

[ \frac{1}{2\sqrt{3x}} \cdot 3 + \frac{1}{2\sqrt{3y^4}} \cdot 12y^3 \cdot \frac{{dy}}{{dx}} = 0 ]

Simplify the equation:

[ \frac{1}{{2\sqrt{3x}}} + \frac{{6y^3}}{{2\sqrt{3}y^2}}\frac{{dy}}{{dx}} = 0 ]

Now, solve for (\frac{{dy}}{{dx}}):

[ \frac{{dy}}{{dx}} = -\frac{1}{{6y^3}} \cdot \frac{{\sqrt{3x}}}{{\sqrt{3}}} ]

[ \frac{{dy}}{{dx}} = -\frac{1}{{6\sqrt{3}y^3}} \cdot \sqrt{x} ]

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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.

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.

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.

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.

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