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Question Video: Finding the Second Derivative of a Function Mathematics • Higher Education

Given that 𝑦 = βˆ’4π‘₯Β²(2π‘₯ βˆ’ 1)Β², find 𝑦″.

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Video Transcript

Given that 𝑦 is equal to negative four π‘₯ squared times two π‘₯ minus one all squared, find 𝑦 double prime.

We’re given 𝑦 as a function in π‘₯. And we’re asked to find an expression for 𝑦 double prime. That’s the second derivative of 𝑦 with respect to π‘₯. And in this case, there’s actually a lot of different options we can do. For example, we could notice that 𝑦 is the product of two functions. So we could do this by using the product rule. However, this is not the only method we could use. For example, we could notice our first factor is squared and our second factor is squared. So we could use our laws of exponents to write this as one expression all squared. For example, since four π‘₯ squared is equal to two π‘₯ all squared, by using our laws of exponents, we can rewrite 𝑦 as negative one times two π‘₯ times two π‘₯ minus one all squared.

So instead, we could differentiate this expression with respect to π‘₯. And there’s several different ways we could do this. We could use the chain rule, the general power rule, or we could expand this and then use the power rule for differentiation. And all of these methods will work. What we’re going to do is distribute the square over the parentheses in our expression for 𝑦. And we can do this by using either binomial expansion or the FOIL method. Either way, we get 𝑦 is equal to negative four π‘₯ times four π‘₯ squared minus four π‘₯ plus one.

Next, we’ll distribute negative four π‘₯ squared over our parentheses. So we need to multiply each term inside of our parentheses by negative four π‘₯ squared. Doing this and then simplifying, we get negative 16π‘₯ to the fourth power plus 16π‘₯ cubed minus four π‘₯ squared. And we can see this is a polynomial. So we can differentiate this with respect to π‘₯ term by term by using the power rule for differentiation. In each term, we want to multiply by our exponent of π‘₯ and then reduce this exponent by one.

Differentiating our first term, we get negative 16 times four π‘₯ cubed. Differentiating our second term, we get 16 times three π‘₯ squared. And differentiating our third term, we get negative four times two π‘₯ to the first power. We can then simplify this expression to get 𝑦 prime is equal to negative 64π‘₯ cubed plus 48π‘₯ squared minus eight π‘₯. However, remember, the question is asking us to find an expression for 𝑦 double prime. So we need to differentiate 𝑦 with respect to π‘₯ twice.

So to do this, we’re going to need to differentiate our expression for 𝑦 prime with respect to π‘₯. Since this is a polynomial, we can do this term by term by using the power rule for differentiation. Differentiating our first term with respect to π‘₯, we get negative 64 times three π‘₯ squared. Differentiating our second term, we get 48 times two π‘₯ to the first power. And differentiating our third term, we get negative eight times one π‘₯ to the zeroth power.

Now all that’s left to do is simplify. In our first term, negative 64 times three is negative 192. In our second term, 48 times two is equal to 96 and π‘₯ to the first power is just equal to π‘₯. And in our third term, negative eight times one is equal to negative eight and π‘₯ to the zeroth power is just equal to one. And this gives us our final answer.

Therefore, we were able to show if 𝑦 is equal to negative four π‘₯ squared times two π‘₯ minus one all squared, then 𝑦 double prime will be equal to negative 192π‘₯ squared plus 96π‘₯ minus eight.

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