We're given "F of x" equals 2 "x to the fourth" minus 5 "e to the x." We want to find the first
derivative function "f prime of x," then "f prime of 1," then the second derivative function or
"f double prime of x," and then "f double prime of 1." So we'll begin by finding "f prime of x,"
which is equal to 2 times the derivative of "x to the fourth" minus 5 times the derivative of "e
to the x." Well, the derivative of "2 x to the fourth," we'd multiply by 4, that would be 8,
subtract one from the exponent, we'd have 8 "x cubed" minus 5 times the derivative of "e to the
x." Well, the derivative of "e to the x" is "e to the x," so we just have minus 5 "e to the x."
And now to find "f prime of 1," we substitute 1 for x, so we'd have 8 times 1 cubed minus 5
"e to the 1." Well, "1 cubed" is 1, 8 times 1 is 8, and then for 5 "e to the 1," we can
make this exponent positive by moving it to the denominator, so this would be the
same as "minus 5 divided by e." Let's also get our decimal approximation for this value. We'd
have 8 minus 5 divided by "e." On this calculator, if we press "second" then "divide," it brings up
the "e," and then "enter." If we round to three decimal places, notice how the
three in the fourth decimal place tells us to round down. This would be approximately 9.839.
Now, we want to find the second derivative function, which should be the derivative of the
first derivative function. So we want to find the derivative of "8 x cubed" minus "5 e to the x."
Well, the derivative of "8 x cubed" would be 24 "x squared," and the derivative of
"e to the x" is "e to the x," so we just have minus 5 "e to the x."
And now to find "f double prime of 1," we substitute 1 for x, which would give us
"positive 1 times 24," so 24 minus "5 divided by e." Let's get a decimal approximation for this
as well. So we have 24 minus "5 divided by e." Again, to three decimal places, notice here we
have a six in the fourth decimal place, so we round up. This would be approximately 22.116.
Now let's look at the graph of our function and at the point on the function when "x equals 1."
The original function is graphed here in blue. Notice when "x equals 1," this
would be the point on the function. To find the corresponding y-coordinate,
we'd have to substitute 1 into the original function. Notice that for this x-value, the first
derivative is approximately 9.839. This tells us the rate of change of the function at this point
and also the slope of the tangent line at this point. So, the slope of this red tangent line
is exactly "8 minus 5 divided by e," which is approximately 9.839. And also because the
first derivative is positive at this point, the function is increasing at this point.
Notice the second derivative, or "f double prime of 1," is approximately 22.116. This tells us the
rate of change of the slope of the tangent line at this point, or we can say the rate of change of
the instantaneous rate of change of the function at "x equals 1." And because it's positive at
this point, the slope of the tangent line is actually increasing, meaning as x increases,
the slope of this tangent line would increase. It also tells us something about the concavity of the
function at this point. And because it's positive, the function is concave up at this point. But
we'll talk more about increasing, decreasing, concave up, concave down in future lessons.
I hope you found this helpful.