Find d^2y/dx^2 if 5x^3 - y^3 = 25

A. -250x/y^5

B. (10x(y^2-x))/y^3

C. 5x^2/y^2

D. (10(y^2-5x^4))/y^3

E. 30x - 6y

Answer:

[tex]y''=\frac{-250x}{y^5}[/tex]

Choice A.

Step-by-step explanation:

We are given:

[tex]5x^3-y^3=25[/tex]

We are asked to find:

[tex]\frac{d^2y}{dx^2}[/tex]

This requires we first find:

[tex]\frac{dy}{dx}[/tex]

Let's begin.

We will differentiate the right hand side using the constant rule since 25 is just a constant. We will differentiate the left hand side using difference rule  and then constant multiple rule and then power rule and chain rule.

[tex](5x^3-y^3=25)'[/tex]

[tex](5x^3-y^3)'=(25)'[/tex]

[tex](5x^3)'-(y^3)'=0[/tex]

[tex]5(x^3)'-(y^3)'=0[/tex]

[tex]15x^2-3y^2y'=0[/tex]

Now let's solve for [tex]y'[/tex].

Add [tex]3y^2y'[/tex] on both sides:

[tex]15x^2=3y^2y'[/tex]

Divide both sides by [tex]3y^2[/tex]:

[tex]\frac{15x^2}{3y^2}=y'[/tex]

Symmetric property of equality:

[tex]y'=\frac{15x^2}{3y^2}[/tex]

Simplify right hand side:

[tex]y'=\frac{5x^2}{y^2}[/tex]

Now let's find the second derivative by differentiating both sides of this equation. We will use the quotient rule on the right hand side:

[tex]y''=\frac{(5x^2)'y^2-5x^2(y^2)'}{(y^2)^2}[/tex]

[tex]y''=\frac{10xy^2-5x^2(2yy')}{y^4}[/tex]

We can replace [tex]y'[/tex] with [tex]\frac{5x^2}{y^2}[/tex]... since we just got this result before differentiating both sides of an equality again.

So we have:

[tex]y''=\frac{10xy^2-5x^2(2y\frac{5x^2}{y^2})}{y^4}[/tex]

Let's simplify the 2nd term on top by cancelling a common factor of y in that term from top and bottom:

[tex]y''=\frac{10xy^2-5x^2(2\frac{5x^2}{y})}{y^4}[/tex]

We still don't like having this compound fraction (a mini fraction inside a bigger fraction). Let's get rid of that...Multiply numerator and denominator by [tex]\frac{y}{y}[/tex] for the big fraction to clear the mini fraction out.

[tex]y''=\frac{y}{y}\frac{10xy^2-5x^2(2\frac{5x^2}{y})}{y^4}[/tex]

Distribute:

[tex]y''=\frac{10xy^3-5x^2(2)5x^2}{y^5}[/tex]

Simplifying a bit more:

[tex]y''=\frac{10xy^3-50x^4}{y^5}[/tex]

We could factor the top a bit... both terms on top of a common factor of [tex]10x[/tex]:

[tex]y''=\frac{10x(y^3-5x^3)}{y^5}[/tex]

Notice the thing in ( ) on top is the exact opposite of the left hand side of our original equation: [tex]5x^3-y^3=25[/tex].

This implies we can replace it with -25:

[tex]y''=\frac{10x(-25)}{y^5}[/tex]

[tex]y''=\frac{-250x}{y^5}[/tex]

f(x)=6x+5 and g(x)=2x-4

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