Find d²y/dx² in terms of x and y.
y^5 = x^7
d²y/dx² =????​

Find Dy/dx In Terms Of X And Y.y^5 = X^7dy/dx =????

Answers

Answer 1

Answer:   [tex]\frac{14y}{25x^2}[/tex]

====================================================

Work Shown:

First calculate [tex]\frac{dy}{dx}[/tex] through the use of implicit differentiation.

Don't forget about the chain rule.

[tex]y^5 = x^7\\\\\frac{d}{dx}\left[y^5\right] = \frac{d}{dx}\left[x^7\right]\\\\5y^4\frac{dy}{dx} = 7x^6\\\\\frac{dy}{dx} = \frac{7x^6}{5y^4}\\\\[/tex]

Go back to line 3, shown above, and apply the derivative to both sides.

You'll be using the product rule.

[tex]5y^4\frac{dy}{dx} = 7x^6\\\\\frac{d}{dx}\left[5y^4\frac{dy}{dx}\right] = \frac{d}{dx}\left[7x^6\right]\\\\\frac{d}{dx}\left[5y^4\right]*\frac{dy}{dx}+5y^4*\frac{d}{dx}\left[\frac{dy}{dx}\right] = 42x^5\\\\20y^3*\frac{dy}{dx}*\frac{dy}{dx}+5y^4*\frac{d^2y}{dx^2}=42x^5\\\\20y^3*\left(\frac{dy}{dx}\right)^2+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\[/tex]

Use substitution and isolate [tex]\frac{d^2y}{dx^2}[/tex] to get the following:

[tex]20y^3*\left(\frac{dy}{dx}\right)^2+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\20y^3*\left(\frac{7x^6}{5y^4}\right)^2+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\\frac{196x^{12}}{5y^5}+5y^4*\frac{d^2y}{dx^2} = 42x^5\\\\\frac{196x^{12}}{5x^7}+5y^4\frac{d^2y}{dx^2} = 42x^5\\\\[/tex]

[tex]\frac{196x^5}{5}+5y^4*\frac{d^2y}{dx^2}=42x^5\\\\5y^4*\frac{d^2y}{dx^2}=42x^5-\frac{196x^5}{5}\\\\5y^4*\frac{d^2y}{dx^2}=\frac{210x^5-196x^5}{5}\\\\5y^4*\frac{d^2y}{dx^2}=\frac{14x^5}{5}\\\\[/tex]

[tex]\frac{d^2y}{dx^2}=\frac{14x^5}{5}*\frac{1}{5y^4}\\\\\frac{d^2y}{dx^2}=\frac{14x^5}{25y^4}\\\\\frac{d^2y}{dx^2}=\frac{14x^5*x^2}{25y^4*x^2}\\\\\frac{d^2y}{dx^2}=\frac{14x^7}{25y^4*x^2}\\\\\frac{d^2y}{dx^2}=\frac{14y^5}{25y^4*x^2}\\\\\frac{d^2y}{dx^2}=\frac{14y}{25x^2}\\\\[/tex]


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Step-by-step explanation:

1 Evaluate the expression 2a + b - 3 when a=5 and b=2. Please click on "show your work" to set up the problem and show your work. Then, enter your answer in the box to the right. You must complete both of these steps. Evaluate the expression 2a + b - 3 when a=5 and b=2. Set up the expression and show your work.

Answers

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Answer:

How does knowing one linear factor of a polynomial help find the other factors?

Step-by-step explanation:

f(x)=(x−3)(x−1)(x+2)(x+6)

f(x)=(x−2)(x−2)(x+3)(x+5)

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f(x)=(x−2)(x−2)(x+3)(x+5)

Explanation:

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The leading coefficient is the number in front of the largest power of the variable.  When the terms are listed in descending order (highest to lowest power), the leading coefficient is always the first number.  In our case the leading coefficient is hard to spot.  Since there is no number in front of x4, the coefficient is 1 by default.

This is nice because the only factor of 1 is well ... 1.

We then create all the possible fractions with a factor of the constant in the numerator and a factor of the leading coefficient in the denominator.  This actually isn't as bad as it could be since our only possible denominator is 1.  Any fraction with a denominator of 1 is just the numerator.  Therfore, our possible "fractions" are simply

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2 does in fact work twice and is thus a double root.  Since we only have three terms remainng, we can convert from synthetic back to an algebraic expression.

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f(x)=(x+3)(x+5)

Writing our root of 2 as an algebraic expression gives (x−2).  Since we have double root, we need two of these.  Therfore, our final factored expression is.

f(x)=(x−2)(x−2)(x+3)(x+5)

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