Question

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 1

Answer:

$y''=\frac{-250x}{y^5}$

Choice A.

Step-by-step explanation:

We are given:

$5x^3-y^3=25$

We are asked to find:

$\frac{d^2y}{dx^2}$

This requires we first find:

$\frac{dy}{dx}$

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.

$(5x^3-y^3=25)'$

$(5x^3-y^3)'=(25)'$

$(5x^3)'-(y^3)'=0$

$5(x^3)'-(y^3)'=0$

$15x^2-3y^2y'=0$

Now let's solve for $y'$.

Add $3y^2y'$ on both sides:

$15x^2=3y^2y'$

Divide both sides by $3y^2$:

$\frac{15x^2}{3y^2}=y'$

Symmetric property of equality:

$y'=\frac{15x^2}{3y^2}$

Simplify right hand side:

$y'=\frac{5x^2}{y^2}$

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:

$y''=\frac{(5x^2)'y^2-5x^2(y^2)'}{(y^2)^2}$

$y''=\frac{10xy^2-5x^2(2yy')}{y^4}$

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

So we have:

$y''=\frac{10xy^2-5x^2(2y\frac{5x^2}{y^2})}{y^4}$

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

$y''=\frac{10xy^2-5x^2(2\frac{5x^2}{y})}{y^4}$

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 $\frac{y}{y}$ for the big fraction to clear the mini fraction out.

$y''=\frac{y}{y}\frac{10xy^2-5x^2(2\frac{5x^2}{y})}{y^4}$

Distribute:

$y''=\frac{10xy^3-5x^2(2)5x^2}{y^5}$

Simplifying a bit more:

$y''=\frac{10xy^3-50x^4}{y^5}$

We could factor the top a bit... both terms on top of a common factor of $10x$:

$y''=\frac{10x(y^3-5x^3)}{y^5}$

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

This implies we can replace it with -25:

$y''=\frac{10x(-25)}{y^5}$

$y''=\frac{-250x}{y^5}$