The problem was to use logarithmic differentiation or the shortcut method to find the derivative of
x2e-xsin(2x)
f(x) =
1 + x2
Again we have the product of four functions:
g1(x) = x2 g1'(x) = 2x
g2(x) = e-x g2'(x) = -e-x
g3(x) = sin(2x) g3'(x) = 2 cos(2x)
1 -2x
g4(x) = g4'(x) =
1 + x2 (1 + x2)2
Step 1: Take the natural log of both sides.
æ x2e-xsin(2x) ö
ln(f(x)) = lnç ÷
è 1 + x2 ø
Step 2: Apply the log identity.
ln(f(x)) = ln(x2) + ln(e-x) + ln(sin(2x)) - ln(1 + x2)which you can simplify further to
ln(f(x)) = 2 ln(x) - x + ln(sin(2x)) - ln(1 + x2)
Step 3: Use the chain rule to take the derivative of both sides.
f'(x) 2 2 cos(2x) 2x=- 1 +-f(x) x sin(2x) 1 + x2
Step 4: Multiply through by f(x).
æ 2 2 cos(2x) 2x ö
f'(x) = ç - 1 + - ÷ f(x)
è x sin(2x) 1 + x2 ø
Step 5: Substitute the original product for f(x).
æ 2 2 cos(2x) 2x ö x2e-xsin(2x)
f'(x) = ç - 1 + - ÷
è x sin(2x) 1 + x2 ø 1 + x2
Step 6 (optional): Multiply it through.
2xe-xsin(2x) x2e-xsin(2x)
f'(x) = - +
1 + x2 1 + x2
2x2e-xcos(2x) 2x3e-xsin(2x)
-
1 + x2 (1 + x2)2
If you used the shortcut method you would have arrived at this point in one
step.
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