What's the derivative of the following function?
However, in order to do this we need to find the derivatives of the numerator and the denominator, each of which requires the product rule.
Now for the quotient rule:
We want to simplify as much as is reasonable, but not too much. In this case the denominator isn't too bad to multiply out, so we'll do that:
Notice that ex is common to every term in the numerator, and is also in the denominator:
We can cancel out ex. We will still have one occurrence of ex in the denominator, since the denominator previously contained (ex)2:
We'll stop here, because if we multiply stuff out we'll be doing more unnecessary steps.
What is the derivative of the following function?
Now we need to use the quotient rule and the product rule only once each, which is much easier. We need the product rule to find the derivative of the denominator:
Again, every term has at least one ex, so we can cancel those out:
In this case, it makes sense to multiply out the terms because there aren't many to multiply out, and because x2 and just multiply out to x.
Find the derivative of the function.
This one has a quotient inside a quotient and there's not much we can do about it. We'll need the derivative of the numerator for use in the quotient rule.
Now for the outer quotient:
It would be reasonable to leave the answer like this. It would also be reasonable to distribute the :
What is f ' (x) for the following function?
Now we stick everything in the product rule:
And we leave it there, because that's too messy to do anything useful with.
Rewrite the function
We can rewrite the function as
f(x) = ex(x2 + 4x)cos x,
which has three factors instead of four. This is definitely an improvement. Since we have three factors we'll use the product rule twice. Break up the function as
f(x) = (ex(x2 + 4x))(cos x).
Then the outermost application of the product rule says that
f ' (x) = (ex(x2 + 4x))'(cos x) + (ex(x2 + 4x))(cos x)'.
We need to use the product rule again to find one of those derivatives:
(ex(x2 + 4x))' = (ex)'(x2 + 4x) + ex(x2 + 4x)'
= ex(x2 + 4x) + ex(2x + 4).
This can be simplified by factoring out ex:
= ex(x2 + 4x + 2x + 4)
= ex(x2 + 6x + 4)
Now go back to the outermost application of the product rule.
f ' (x) = (ex(x2 + 4x))'(cos x) + (ex(x2 + 4x))(cos x)'
= (ex(x2 + 6x + 4))cos x + (ex(x2 + 4x))(-sin x)
= (ex(x2 + 6x + 4))cos x – (ex(x2 + 4x))(sin x)
Since each term has a factor of ex, we'll factor that out:
f ' (x) = ex[(x2 + 6x + 4)cos x – (x2 + 4x)(sin x)]
And now we're done.
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