Video Transcript
Given that 𝑦 equals 𝑎𝑥 cubed
plus 𝑏𝑥 squared, the third-order derivative of 𝑦 is equal to negative 18, and 𝑑
two 𝑦 over 𝑑𝑥 squared when 𝑥 is equal to two is equal to negative 14, find 𝑎
and 𝑏.
So the first thing we need to do in
this question is actually find our first-, second-, and third-order derivatives. So we’re gonna start with our
first-order derivative. And just to remind us how we
actually differentiate if we have a function in 𝑥, then if it’s in the form 𝑎𝑥 to
the power of 𝑏, then our first-order derivative is gonna be equal to 𝑎𝑏𝑥 to the
power of 𝑏 minus one. So you multiply the coefficient by
the exponent and reduce the exponent by one.
Okay, so let’s use that and let’s
find the first-order derivative of our function. So the first term is gonna be three
𝑎𝑥 squared. And that’s because we multiplied
our exponent three by a coefficient 𝑎. So we get three 𝑎 and then we’ve
reduced the exponent by one. So we get 𝑥 squared. So then, the second term is just
plus two 𝑏𝑥. Again, we’ve differentiated that
term. Okay, so that’s our first-order
derivative.
What we’re gonna move onto now is
going to differentiate again to find our second-order derivative. So our second-order derivative
which I’ve shown here with 𝑦 prime prime. And all as we’ve got it in the
question 𝑑 two 𝑦 over 𝑑𝑥 squared is gonna be six 𝑎𝑥 plus two 𝑏. So again, we just differentiate
that in the normal way. So we have three 𝑎 multiplied by
two — our coefficient multiplied by our exponent — and then 𝑥 to the power of two
minus one, which is just 𝑥 — so 𝑥 to the power of one. Okay, great, so now we’ve found our
second-order derivative.
So now, great, we can actually move
on and find our third-order derivative. And all we do to find this is
actually differentiate again, which is just gonna leave us with six 𝑎 because if we
differentiate six 𝑎𝑥, we’re just left with six 𝑎. And if we differentiate plus two
𝑏, then it’s just going to be zero because if we differentiate any number that
hasn’t got an 𝑥 term in, it just goes to zero. Okay, so now we’ve done that —
we’ve got our first-, second-, and third-order derivatives — let’s see what we do
next.
Well, let’s start with this bit of
information. We know that our third-order
derivative is gonna be equal to negative 18. So we can now set up an
equation. And we can actually set up that new
equation because we can say that our third-order derivative is equal to negative
18. So can sub that in. So we get negative 18 is equal to
six 𝑎. Okay, so therefore, we can say that
𝑎 is going to be equal to negative three. And we’ve got that because we’ve
actually divided through by six. And also, we just flipped it just
so we got the 𝑎 on the left-hand side.
So fantastic, we’ve now found
𝑎. So now, what we need to do is move
on and have a go at finding 𝑏. And in order to do that, what we’re
gonna use is this bit of information here. Because if we take a look at this
bit of information, we’ve got that 𝑑 two 𝑦 over 𝑑𝑥 squared — so our second-order
derivative — is equal to negative 14 when 𝑥 is equal to two. So therefore, in order to actually
find 𝑏, what we can do is substitute in 𝑥 is equal to two and 𝑦 prime prime is
equal to negative 14. And this is gonna give us that
negative 14 is equal to six 𝑎 multiplied by two plus two 𝑏.
But we’ve now got something
else. We’re gonna substitute in because
we know 𝑎. So we can substitute in 𝑎 is equal
to negative three. So therefore, we get that negative
14 equals six multiplied by negative three multiplied by two plus two 𝑏. So we have negative 14 equals
negative 36 plus two 𝑏. So then, what we do is we actually
add 36 to each side. So we have 22 equals two 𝑏. So therefore, if we divide by two,
we get 𝑏 is equal to 11.
So that means that we can say that
given that 𝑦 equals 𝑎𝑥 cubed plus 𝑏𝑥 squared which third-order derivative of 𝑦
is equal to negative 18 and 𝑑 two 𝑦 over 𝑑𝑥 squared when 𝑥 is equal to two is
equal to negative 14, 𝑎 is equal to negative three and 𝑏 is equal to 11.
