Video Transcript
Find the values of 𝑥 for which d𝑦 by d𝑥 is equal to zero, where 𝑦 is equal to 𝑥 squared plus six 𝑥 plus 36 all over 𝑥 squared minus six 𝑥 plus 36.
We’re given a rational function 𝑦 which is a function of 𝑥. And we’re asked to find the values of 𝑥 for which d𝑦 by d𝑥 is equal to zero. The first thing we need to do then is to differentiate 𝑦 with respect to 𝑥 to find d𝑦 by d𝑥. Since our function 𝑦 is a rational function, that is, a fraction or a quotient, we can use the quotient rule for differentiation. Using the notation d𝑓 by d𝑥 is equal to 𝑓 prime, the quotient rule says that for 𝑓 of 𝑥 equal to 𝑢 of 𝑥 over 𝑣 of 𝑥, where 𝑢 and 𝑣 are differentiable functions of 𝑥, 𝑓 prime of 𝑥 is equal to 𝑣 of 𝑥 times 𝑢 prime of 𝑥 minus 𝑢 of 𝑥 times 𝑣 prime of 𝑥 all over 𝑣 of 𝑥 squared.
For our function 𝑦 then, with 𝑢 of 𝑥 equal to 𝑥 squared plus six 𝑥 plus 36 and 𝑣 of 𝑥 equal to 𝑥 squared minus six 𝑥 plus 36, to use the quotient rule, we’re going to need to find 𝑢 prime of 𝑥 and 𝑣 prime of 𝑥. And since both 𝑢 and 𝑣 are polynomials in 𝑥, we can use the power rule for differentiation. This says that for a function 𝑎𝑥 raised to the 𝑛th power where 𝑎 and 𝑛 are constants, its derivative with respect to 𝑥 is 𝑛𝑎𝑥 raised to the 𝑛 minus oneth power. That is, we multiply by the exponent of 𝑥 and subtract one from the exponent.
We can differentiate both 𝑢 and 𝑣 term by term since the derivative of the sum is the sum of the derivatives. If we look at our first term in 𝑢, we have 𝑥 squared. And with the exponent two, we subtract one from the exponent and multiply by the original exponent two. This gives us two times 𝑥 raised to the power two minus one, that is, two times 𝑥 raised to the power one. And since anything raised to the power one is itself, that’s two 𝑥, so the derivative of 𝑥 squared is two 𝑥. And since we have the same first term with 𝑣, then that derivative is also two 𝑥.
Our second term in 𝑢 is six 𝑥. To differentiate this, we can take our six to the front since it’s a constant. And we know that our exponent of 𝑥 is actually one. The derivative with respect to 𝑥 is therefore six times one, the exponent, times 𝑥 raised to the power one minus one. That is six 𝑥 raised to the power zero. And we know that anything to the power zero is equal to one so that the derivative of six 𝑥 with respect to 𝑥 is six. We can apply exactly the same process to 𝑣 where the second term is negative six, where in this case the derivative with respect to 𝑥 is negative six.
Our final term in both 𝑢 and 𝑣 is 36. This is a constant which does not depend on 𝑥. Therefore, the derivative with respect to 𝑥 is equal to zero. And so we have 𝑢 prime of 𝑥 is two 𝑥 plus six and 𝑣 prime of 𝑥 is two 𝑥 minus six. So now we can apply the quotient rule to our function 𝑦. We have 𝑣 of 𝑥, which is 𝑥 squared minus six 𝑥 plus 36, times 𝑢 prime of 𝑥, which is two 𝑥 plus six, minus 𝑢 of 𝑥, which is 𝑥 squared plus six 𝑥 plus 36, times 𝑣 prime of 𝑥, which is two 𝑥 minus six. And that’s all over 𝑣 of 𝑥 squared, which is 𝑥 squared minus six 𝑥 plus 36 all squared.
So now distributing our parentheses, in the numerator, we have two 𝑥 cubed minus 12𝑥 squared plus 72𝑥 plus six 𝑥 squared minus 36𝑥 plus 216 minus two 𝑥 cubed plus 12𝑥 squared plus 72𝑥 minus six 𝑥 squared minus 36𝑥 minus 216 all over 𝑥 squared minus six 𝑥 plus 36 all squared. We can collect our like terms in each of our parentheses. So we have minus 12𝑥 squared plus six 𝑥 squared in the first set of parentheses and 12𝑥 squared minus six 𝑥 squared in the second. And we have 72𝑥 minus 36𝑥 in the first and 72𝑥 minus 36𝑥 in the second, which gives us two 𝑥 cubed minus six 𝑥 squared plus 36𝑥 plus 216 minus two 𝑥 cubed plus six 𝑥 squared plus 36𝑥 minus 216 all over 𝑥 squared minus six 𝑥 plus 36 all squared.
Now, distributing the negative sign across our second set of parentheses, we can cancel the two 𝑥 cubed with the negative two 𝑥 cubed, the 36𝑥 with the negative 36𝑥. Making some room and collecting like terms, we then have d𝑦 by d𝑥 is equal to negative 12𝑥 squared plus 432 all over 𝑥 squared minus six 𝑥 plus 36 squared. So this is d𝑦 by d𝑥, but we’re asked in the question to find the values of 𝑥 for which d𝑦 by d𝑥 is equal to zero. And we can begin to solve this by taking the common factor of negative 12 in our numerator outside some parentheses.
So now we need to solve negative 12 times 𝑥 squared minus 36 all over 𝑥 squared minus six 𝑥 plus 36 squared is equal to zero. It’s perhaps worth noting that our denominator has no real roots. That is, there are no real values of 𝑥 which make the denominator equal to zero. And so this expression is never undefined. So now, to solve this for 𝑥, if we multiply both sides by the denominator, on the left-hand side, we can cancel and on the right-hand side, anything multiplied by zero is zero. So now we simply need to solve negative 12 times 𝑥 squared minus 36 is equal to zero.
If we divide both sides through by negative 12, on the left-hand side, the negative 12s cancel each other out. And on the right-hand side, zero divided by any nonzero integer is equal to zero. And so it remains to solve 𝑥 squared minus 36 is equal to zero. Adding 36 to both sides gives us 𝑥 squared equal to 36. And taking square roots on both sides gives us 𝑥 equal to plus or minus six so that the values of 𝑥 for which d𝑦 by d𝑥 is equal to zero where 𝑦 is equal to 𝑥 squared plus six 𝑥 plus 36 over 𝑥 squared minus six 𝑥 plus 36 are 𝑥 equal to plus or minus six.
