© 2002 by Karl Hahn
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Problem 6 was to find some difficult derivatives:
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i) f(x) = cos(x)√sin2(x) + 1
Your hint was that you will need to apply the product rule once and the
chain rule twice. The two functions to which you apply the product rule are
g(x) = cos(x)
and
___________
h(x) = √sin2(x) + 1
Since f(x) = g(x)h(x), the
product rule says that
f'(x) = g'(x)h(x) + h'(x)g(x)You already should know that g'(x) = -sin(x). But finding the derivative of h(x) is a little harder. Observe though that h(x) is a composite of h(x) = u(v(x)), where
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u(v) = √v
and
v(x) = sin2(x) + 1We know from the power rule (or from the table of rules for taking derivatives) that
1 1
u'(v) = _
2 √v
and we know from the chain rule that if
h(x) = u(v(x)), then
h'(x) = u'(v(x))v'(x)We can already fill in the expression for u'(v(x)) based upon what we have already done:
1 1
h'(x) = ___________ v'(x)
2 √sin2(x) + 1
So the only problem left is finding v'(x). Observe that v(x)
contains a composite function itself -- that is the function,
sin2(x). In other words, it's a composite of taking
the sine of x and squaring the result. If you apply the chain rule
to that, you find that the derivative of sin2(x)
is 2sin(x)cos(x). Make sure you see why.
Since the derivative of the constant, 1, is zero, this expression
is also the derivative of
v(x) = sin2(x) + 1.
So now you know what to put in for v'(x).
1 1
h'(x) = ___________ (2 sin(x)cos(x))
2 √sin2(x) + 1
Now that we know g'(x) and h'(x), we are ready to write
the derivative (using the product rule) of f(x) = g(x)h(x).
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f'(x) = -sin(x)√sin2(x) + 1 +
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1 cos(x) |
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f'(x) = -sin(x)√sin2(x) + 1 +
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sin(x)cos2(x) |
cos(x2)
ii) f(x) =
x2 - 2x + 14
Your clue here was to use the quotient rule
and the chain rule. You have
f(x) = g(x)/h(x), where
g(x) = cos(x2), and
h(x) = x2 - 2x + 14.
You should quickly be able to find that
h'(x) = 2x - 2But g'(x) is not quite as easy to find. Observe that g(x) is a composite of squaring x, and then taking the cosine of the result. So you have g(x) = u(v(x)), where u(v) = cos(v) and v(x) = x2. The chain rule says that
g'(x) = u'(v(x))v'(x)You ought to be able to find for yourself that u'(v) = -sin(v) and v'(x) = 2x. Putting those into the above equation for g'(x) you get:
g'(x) = -sin(x2)(2x)Now that we know g'(x) and h'(x), we are ready to put them into the quotient-rule formula:
h(x)g'(x) - g(x)h'(x)
f'(x) = =
h2(x)
(x2 - 2x + 14)(-sin(x2)(2x)) - cos(x2)(2x - 2)
(x2 - 2x + 14)2
Your instructor might want you to algebraically munge this into some other
(presumably simpler) form, but as far as I'm concerned, the above is a
satisfactory answer.
On the next one, find the second derivative of the function given:
iii) f(x) = e-x sin(1/x)
The hint says to use the product rule and the chain rule to first take the first derivative, and then take the derivative of that to get the second derivative. Clearly you have f(x) = g(x)h(x), where g(x) = e-x and h(x) = sin(1/x). You should already know that g'(x) = -e-x. To find the derivative of h(x), you have you use the chain rule. Observe that h(x) = u(v(x)), where u(v) = sin(v) and v(x) = 1/x. You should know from this that u'(v) = cos(v) and v'(x) = -1/x2. The chain rule says that
cos(1/x)
h'(x) = u'(v(x))v'(x) = - |
-cos(1/x)(x-2) |
f'(x) = g'(x)h(x) + h'(x)g(x) = -e-x sin(1/x) - e-x cos(1/x)(x-2)This, of course, is only the first derivative. We have to take the derivative of this expression to find the second derivative of the original function. Fortunately the first summand is a very near copy of the original function -- in fact it is the negative of the original function. So finding its derivative is already done. The second summand is the product of three functions: p(x) = e-x, q(x) = cos(1/x), and r(x) = x-2. To apply the product rule to a product of three factors, you simply apply it twice:
(p(x)q(x)r(x))' = p'(x)(q(x)r(x)) + p(x)(q(x)r(x))'
= p'(x)q(x)r(x) + p(x)(q'(x)r(x) + r'(x)q(x))
= p'(x)q(x)r(x) + q'(x)p(x)r(x) + r'(x)p(x)q(x)
We already know that
p'(x) = -e-x.
By analogy with sin(1/x) we can easily see that
q'(x) = sin(1/x)(x-2).
And by the power rule,
r'(x) = -2x-3.
So
(p(x)q(x)r(x))' =
-e-x cos(1/x)(x-2) + e-x sin(1/x)(x-4) - 2e-x cos(1/x)(x-3)
Putting all the pieces together (and you should try to do this yourself without
looking at the final answer), we have
f"(x) = e-x sin(1/x) + e-x cos(1/x)(x-2) +
e-x cos(1/x)(x-2) - e-xsin (1/x)(x-4) + 2e-x cos(1/x)(x-3)
Notice that the second and third terms are the same. So this simplifies to
f"(x) = e-x sin(1/x) + 2e-x cos(1/x)(x-2) -
e-x sin(1/x)(x-4) + 2e-x cos(1/x)(x-3)
h'(x) = -cos(1/x)(x-2)
h"(x) = -sin(1/x)(x-2)(x-2) + 2cos(1/x)(x-3)
= -sin(1/x)(x-4) + 2cos(1/x)(x-3)
Make sure you see where all that came from. Leibniz says to combine them like this:
f"(x) = g"(x)h(x) + 2g'(x)h'(x) + g(x)h"(x)Try it yourself from the expressions we came up with above, and confirm that you get the same answer as by doing it the first way we did it.