Video: Differentiating Twice and Solving an Equation in the Second Derivative to Find the Value of an Unknown Constant

Given 𝑦 = βˆ’π‘₯Β² βˆ’ 8π‘₯ βˆ’ 8 and d²𝑦/dπ‘₯Β² βˆ’ 9π‘˜ + 4 = 8, find the value of π‘˜.

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Video Transcript

Given 𝑦 is equal to negative π‘₯ squared minus eight π‘₯ minus eight and d squared 𝑦 by dπ‘₯ squared minus nine π‘˜ plus four is equal to eight, find the value of π‘˜.

We’re given 𝑦, which is a function of π‘₯, and an expression with the second derivative of 𝑦 with respect to π‘₯ and containing the unknown constant π‘˜. And we’re asked to find the value of π‘˜. We can do this by differentiating our function 𝑦 twice to find d squared 𝑦 by dπ‘₯ squared and then evaluating the given expression with our result to find π‘˜. So with our function 𝑦 is negative π‘₯ squared minus eight π‘₯ minus eight, let’s find the first derivative of 𝑦 with respect to π‘₯ d𝑦 by dπ‘₯. We can differentiate 𝑦 term by term using the power rule for derivatives. This says for a function 𝑦 equal to π‘Žπ‘₯ raised to the 𝑛th power, d𝑦 by dπ‘₯ is equal to π‘›π‘Žπ‘₯ raised to the 𝑛 minus oneth power. That is, we multiply by the exponent and subtract one from the exponent.

In our first term, our exponent is two so that the derivative of our first term is negative two times π‘₯ raised to the power two minus one. Two minus one is one, so we have negative two times π‘₯ raised to the power one, which is negative two π‘₯. For our next term, negative eight π‘₯, that’s actually negative eight π‘₯ raised to the power one. And using the power rule again, we multiply by our exponent and subtract one from the exponent, which gives us negative eight times one times π‘₯ raised to the power one minus one, which is negative eight π‘₯ raised to the power zero. And we know that anything to the power zero is equal to one. So the derivative of our second term is negative eight. And so far we have d𝑦 by dπ‘₯ is negative two π‘₯ minus eight. And since our third term, negative eight, is a constant β€” it doesn’t depend on π‘₯ β€” and so its derivative is zero. Our first derivative, d𝑦 by dπ‘₯, is therefore negative two π‘₯ minus eight.

We’re given an expression involving d squared 𝑦 by dπ‘₯ squared. And to find this, we need to differentiate our first derivative, that is, find d by dπ‘₯ of negative two π‘₯ minus eight. If we differentiate again term by term using the power rule, our first term is negative two π‘₯, which is actually negative two times π‘₯ raised to the power one. And so multiplying by the exponent and subtracting one from the exponent, we have d by dπ‘₯ of negative two π‘₯ is negative two times one times π‘₯ raised to the power one minus one. That is negative two π‘₯ raised to the power zero. And we know that anything to the power zero is equal to one so that the derivative of our first term is negative two.

Our second term is negative eight, again a constant. And since the rate of change of a constant does not depend on π‘₯, its derivative is equal to zero, we have d squared 𝑦 by dπ‘₯ squared is equal to negative two. Now in the question, we’re given d squared 𝑦 by dπ‘₯ squared minus nine π‘˜ plus four is equal to eight. And we want to solve this for π‘˜. And substituting d squared 𝑦 by dπ‘₯ squared is equal to negative two, we have negative two minus nine π‘˜ plus four is equal to eight. Now, if we rearrange this to get the π‘˜β€™s on the right-hand side and the constants on the left, we have negative two plus four minus eight is equal to nine π‘˜.

Evaluating the left-hand side gives us negative six is equal to nine π‘˜. And dividing through by nine, we can cancel the nine on the right-hand side to give us negative six over nine is equal to π‘˜. We have a common factor of three in the numerator and the denominator on the left-hand side. And this gives us negative two over three. And so given 𝑦 is equal to negative π‘₯ squared minus eight π‘₯ minus eight and d squared 𝑦 by dπ‘₯ squared minus nine π‘˜ plus four is equal to eight, we find the value of π‘˜ equal to negative two over three.

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