derivative of sinx ki power cosx
Answers
Answer:
This is such a common occurrence! All of the other answers in this thread show no explanation and a bunch of work with no telling of where it came from! I will try to explain this in such a way that you can hopefully understand.
What is the answer to ddx[sincosxx] ?
Let’s get started by setting this mess equal to a function called y , since I’m so original.
y=sincosxx
What I’d do from here is take the natural log of both sides, like this:
ln[y]=ln[sincosxx]
There’s a really interesting trick to logarithms that states this:
If you see this ever happen:
logaxy
Then you can do this with it:
ylogax
In other words, if the argument of the logarithm has an exponent, you can bring that exponent down and make it a coefficient of the log. This is very useful and you will be doing this a lot, so don’t forget it. Let’s apply it to the problem we are currently working with:
ln[(sinx)cosx]=cos(x)ln[sinx]
*Yes, I’m aware the function looks different now. They mean the same thing, though, so no worries.
This now looks like some simple chain rule applications to me!
ddx[lny]=ddx[cos(x)ln(sinx)]
Let’s start taking derivatives!
We will be applying the product rule, which is given by:
ddx[f(x)g(x)]=f′(x)g(x)+f(x)g′(x)
ddx[lny]=ddx[cos(x)ln(sinx)]
We will use the Product Rule here:
ddx[cosx]lnsinx+ddx[lnsinx]cosx=ddxlny
Solving for the derivatives brought upon us by the Product Rule:
(−sinx)lnsinx+(1sinx⋅cosx)cosx=1y⋅dydx
As a quick reminder, the derivative of the natural log of anything is:
ddx[lnf(x)]=1f(x)⋅f′(x)
Back to the problem, we have some simplifications to make here:
−sin(x)ln(sinx)+cosxsinx⋅cosx=1y⋅dydx
Next, we can rewrite cosxsinx as cotx . Also, we can multiply both sides by y to isolate the dydx
y(−sinxlnsinx+cotxcosx)=dydx
This is technically an answer, but it is not what the guy asking the question is looking for. If we gave this to him, he’d be nonplussed; he never gave us a y variable to start with! Since we were the ones that introduced it, we must be the ones to remove it.
Luckily, we know the value of y . It was our original function, because that is what we set it equal to. Remember this?
Let’s get started by setting this mess equal to a function called y , since I’m so original.
y=sincosxx
Now, we’ll conveniently replace all of our y 's with sincosxx 's. Once we’re done with that, we’ll be done with the whole problem.
ddx[sincosxx]=sincosxx(−sinxlnsinx+cotxcosx)
And there it is, ladies and gentlemen! Hope you enjoyed the show
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Answer:
The product rule can be used to differentiate any function of the form f(x)=g(x)h(x) . It states that f'(x)=g'(x)h(x)+g(x)h'(x) . The derivative of sinx is cosx and the derivative of cosx is −sinx .
or you could watch this video:
Step-by-step explanation: