Video Transcript
Given that 𝑦 is equal to six
𝑥 to the power of five plus three 𝑥 squared minus seven 𝑥 plus six, determine
the second derivative of 𝑦 with respect to 𝑥.
Here we’ve been given a
function in 𝑥. And we’re being asked to find d
two 𝑦 by d𝑥 squared. To do this, we’ll differentiate
once to find d𝑦 by d𝑥 then differentiate that once again to find the second
derivative. We recall that we can
differentiate a function of the form 𝑎𝑥 to the power of 𝑛 with respect to 𝑥
for some constant rational number 𝑛, which is not equal to zero and some
constant 𝑎. And we get 𝑛𝑎 times 𝑥 to the
power of 𝑛 minus one. In other words, we borrow the
power of 𝑥 and we make it the coefficient of the derivative. And then, we subtract one from
the power.
In this special case where 𝑛
is equal to zero, we actually have a constant. Let’s call that 𝑏. And the derivative of a
constant is zero. This is going to help us find
the first derivative of our function. The derivative of six 𝑥 to the
power of five is going to be five times six 𝑥. And then, we subtract one from
the power. Five minus one is four. That’s 30𝑥 to the fourth
power. We’ll repeat this to
differentiate three 𝑥 squared. It’s going to be two times
three 𝑥. And then, we subtract one from
the power. Two minus one is one. So, the derivative of three 𝑥
squared with respect to 𝑥 is six 𝑥.
The derivative of negative
seven 𝑥 is one times negative seven 𝑥 to the power of zero. Well, that’s just negative
seven. And, of course, six is a
constant, so the derivative of six is zero. d𝑦 by d𝑥 then, the first
derivative of our equation, is 30𝑥 to the power of four plus six 𝑥 minus
seven. We’ll differentiate each part
of this expression once more to find the second derivative.
We’ll do it piece-by-piece. The derivative of 30𝑥 to the
power of four is four times 30𝑥 to the power of three. The derivative of six 𝑥 is
six. And the derivative of negative
seven is zero. Simplifying fully, and we see
that the second derivative of 𝑦 with respect to 𝑥 is 120𝑥 cubed plus six.
In this example, we saw how we
could apply repeated differentiation to help us find the second derivative of a
polynomial function.
