Question Video: Finding the Second Derivative of a Function | Nagwa Question Video: Finding the Second Derivative of a Function | Nagwa

Question Video: Finding the Second Derivative of a Function Mathematics • Third Year of Secondary School

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Given that 𝑦 = (βˆ’4π‘₯ + 7)(βˆ’7π‘₯Β² βˆ’ 4), determine d²𝑦/dπ‘₯Β².

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Video Transcript

Given that 𝑦 is equal to negative four π‘₯ plus seven multiplied by negative seven π‘₯ squared minus four, determine d two 𝑦 by dπ‘₯ squared.

The question tells us that 𝑦 is equal to the product of two polynomials, and it wants us to determine d two 𝑦 by dπ‘₯ squared. And we know this means the second derivative of 𝑦 with respect to π‘₯. We’re going to need to differentiate this expression twice. Since we’re given 𝑦 as the product of two functions, we might be tempted to use the product rule, and this would work. However, since our factors are just polynomials with two terms each, it’s actually simpler in this case to just multiply out our factors.

Since we’ll just get a polynomial with four terms, we’ll multiply these together by using the FOIL method. Let’s start by multiplying the first two terms. We get negative four times negative seven π‘₯ squared is equal to 28π‘₯ cubed. Next, we need to multiply our outer terms. This gives us negative four π‘₯ times negative four, which is equal to 16π‘₯. Next, we want to multiply our inner two terms. This gives us seven multiplied by negative seven π‘₯ squared, which is equal to negative 49π‘₯ squared. Finally, we want to multiply the last two terms in our factors. We get seven times negative four, which is equal to negative 28.

So, we found the following expression for 𝑦. It’s a polynomial with four terms. Let’s switch our middle two terms around so we’re writing it in decreasing exponents of π‘₯. This gives us 𝑦 is equal to 28π‘₯ cubed minus 49π‘₯ squared plus 16π‘₯ minus 28. Remember, the question wants us to find the second derivative of 𝑦 with respect to π‘₯. We’ll start by finding the first derivative of 𝑦 with respect to π‘₯. That’s d𝑦 by dπ‘₯, which is equal to the derivative of 28π‘₯ cubed minus 49π‘₯ squared plus 16π‘₯ minus 28 with respect to π‘₯.

But this is just the derivative of a polynomial. We can do this term by term by using the power rule for differentiation, which tells us for constants π‘Ž and 𝑛, the derivative of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times π‘Ž times π‘₯ to the power of 𝑛 minus one. We multiply by the exponent of π‘₯ and reduce this exponent by one. Using this, we’ll differentiate our first term. We get three times 28 times π‘₯ to the power of three minus one. This simplifies to give us 84π‘₯ squared. We can also use this to differentiate our second term. We get negative 49 times two, which is negative 98. And then, we have π‘₯ to the power of two minus one, which is π‘₯.

We could also differentiate our last two terms by using the power rule for differentiation. However, it’s simpler to notice that these two terms make a linear function. So, their slope will be the coefficient of π‘₯, which is 16. So, we found an expression for d𝑦 by dπ‘₯. We can use this to find an expression for our second derivative of 𝑦 with respect to π‘₯. Our second derivative of 𝑦 with respect to π‘₯ would just be the derivative of d𝑦 by dπ‘₯ with respect to π‘₯.

So, to find d two 𝑦 by dπ‘₯ squared, we’re going to differentiate d𝑦 by dπ‘₯ with respect to π‘₯. That’s the derivative of 84π‘₯ squared minus 98π‘₯ plus 16 with respect to π‘₯. And again, this is the derivative of a polynomial. So, we can do this by using the power rule for differentiation. Our first term will be two times 84 times π‘₯ to the first power, which is 168π‘₯. And our second term will just be the coefficient of π‘₯, which is 98.

Therefore, we’ve shown if 𝑦 is equal to negative four π‘₯ plus seven times negative seven π‘₯ squared minus four, then d two 𝑦 by dπ‘₯ squared is equal to 168π‘₯ minus 98.

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