Question

Please solve it.......​

Answer 1

Answer:

dy/dx = - (y/x)^4

Step-by-step explanation:

We are given the equation:

$\displaystyle \large{ \frac{1}{ {x}^{3} } + \frac{1}{ {y}^{3} } = {a}^{3} }$

Recall important differential formulas such as:

$\displaystyle \large{ \frac{d}{dx} (f(x) + g(x)) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)} \\ \displaystyle \large{ \frac{d}{dx} k = 0 \: \: \tt(k \: \: is \: \: a \: \: constant.)}$

Above is a basic property of differential. Looking at the equation above, differentiate both sides with respect to x. (dy/dx)

$\displaystyle \large{ \frac{d}{dx}( \frac{1}{ {x}^{3} } + \frac{1}{ {y}^{3} } )= \frac{d}{dx} {a}^{3} }$

Apply the distribution property in.

$\displaystyle \large{ \frac{d}{dx} \frac{1}{ {x}^{3} } + \frac{d}{dx} \frac{1}{ {y}^{3} } = \frac{d}{dx} {a}^{3} }$

We treat x^3 and y^3 as function to one another, this type of differential will be called implicit once another variable of function is involved f(x,y).

But we treat a^3 as a constant so we just put 0.

$\displaystyle \large{ \frac{d}{dx} \frac{1}{ {x}^{3} } + \frac{d}{dx} \frac{1}{ {y}^{3} } = 0 }$

Recall the quotient rules for the numerator = 1.

$\displaystyle \large{ \frac{d}{dx} (\frac{1}{f(x)}) = -\frac{ \frac{d}{dx} (f(x))}{ {(f(x))}^{2} } }$

Thus:

$\displaystyle \large{ - \frac{ \frac{d}{dy} ( {x}^{3} )}{ ({x}^{3})^{2} } + \frac{d}{dx} \frac{1}{ {y}^{3} } = 0 } \\ \displaystyle \large{- \frac{ 3 {x}^{2} }{ {x}^{6} } + \frac{d}{dx} \frac{1}{ {y}^{3} } = 0 }$

Make sure to recall power rules for differential and exponent laws.

From 3x^2 / x^6.

$\displaystyle \large{ - \frac{ 3 }{ {x}^{4} } + \frac{d}{dx} \frac{1}{ {y}^{3} } = 0 }$

Now what do we do with 1/y^3? We just differentiate it normally but we multiply with dy/dx once we differentiate the y function.

$\displaystyle \large{- \frac{ 3 }{ {x}^{4} } -\frac{ \frac{d}{dx}( {y}^{3}) }{ ({y}^{3})^{2} } = 0 } \\ \displaystyle \large{ -\frac{ 3 }{ {x}^{4} } - \frac{ 3 {y}^{2} }{ {y}^{6}} = 0 } \\ \displaystyle \large{ - \frac{ 3 }{ {x}^{4} } - \frac{ 3 }{ {y}^{4}} \frac{dy}{dx} = 0 }$

Solve for dy/dx, add both sides by 3/x^4.

$\displaystyle \large{ - \frac{ 3 }{ {x}^{4} } + \frac{3}{ {x}^{4} } - \frac{ 3 }{ {y}^{4}} \frac{dy}{dx} = 0 + \frac{3}{ {x}^{4} } } \\ \displaystyle \large{ - \frac{ 3 }{ {y}^{4}} \frac{dy}{dx} = \frac{3}{ {x}^{4} } }$

Divide both sides by - 3/y^4.

$\displaystyle \large{ \frac{- \frac{3}{ {y}^{4} } \frac{dy}{dx} }{ - \frac{3}{ {y}^{4} } } = \frac{ \frac{3}{ {x}^{4} } }{ - \frac{3}{ {y}^{4} } } } \\ \displaystyle \large{ \frac{dy}{dx} = \frac{3}{ {x}^{4} } \times (-\frac{ {y}^{4} }{3} )} \\ \displaystyle \large{ \frac{dy}{dx} = \frac{1}{ {x}^{4} } \times (- \frac{ {y}^{4} }{1}) } \\ \displaystyle \large{ \frac{dy}{dx} = - \frac{ {y}^{4} }{ {x}^{4} } } \\ \displaystyle \large{ \frac{dy}{dx} = - ( { \frac{y}{x} })^{4}}$