If y = (sin x) cosx . Find dy/dx
Answers
Answer:
[math]\frac{d}{dx}[\sin^{\cos x}x]=\sin^{\cos x}x(-\sin x\ln\sin x + \cot x\cos x) [/math]
Step-by-step explanation:
Let’s get started by setting this mess equal to a function called [math]y[/math], since I’m so original.
[math]y=\sin^{\cos x}x[/math]
What I’d do from here is take the natural log of both sides, like this:
[math]\ln [y]=\ln[\sin^{\cos x}x][/math]
There’s a really interesting trick to logarithms that states this:
If you see this ever happen:
[math]\log_a {x^y}[/math]
Then you can do this with it:
[math]y\log_a x[/math]
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:
[math]\ln[(\sin x)^{\cos x}] = \cos (x)\ln[\sin x][/math]
*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!
[math]\frac{d}{dx}[\ln y] = \frac{d}{dx}[\cos (x)\ln(\sin x)][/math]
Let’s start taking derivatives!
We will be applying the product rule, which is given by:
[math]\frac{d}{dx}[f(x)g(x)] = f'(x)g(x)+f(x)g'(x)[/math]
[math]\frac{d}{dx}[\ln y] = \frac{d}{dx}[\cos (x)\ln(\sin x)][/math]
We will use the Product Rule here:
[math]\frac{d}{dx}[\cos x]\ln\sin x + \frac{d}{dx}[\ln\sin x]\cos x = \frac{d}{dx}\ln y[/math]
Solving for the derivatives brought upon us by the Product Rule:
[math](-\sin x)\ln\sin x + (\frac{1}{\sin x}\cdot\cos x)\cos x = \frac{1}{y}\cdot\frac{dy}{dx}[/math]
As a quick reminder, the derivative of the natural log of anything is:
[math]\frac {d}{dx}[ln f(x)] = \frac{1}{f(x)}\cdot f'(x)[/math]
Back to the problem, we have some simplifications to make here:
[math]-\sin (x)\ln(\sin x)+\frac{\cos x}{\sin x}\cdot\cos x = \frac{1}{y}\cdot\frac{dy}{dx}[/math]
Next, we can rewrite [math]\frac{\cos x}{\sin x}[/math] as [math]\cot x[/math]. Also, we can multiply both sides by [math]y[/math] to isolate the [math]\frac{dy}{dx}[/math]
[math]y(-\sin x\ln\sin x + \cot x\cos x) = \frac{dy}{dx}[/math]
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 [math]y[/math] 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 [math]y[/math]. 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 [math]y[/math], since I’m so original.
[math]y=\sin^{\cos x}x[/math]
Now, we’ll conveniently replace all of our [math]y[/math]'s with [math]\sin^{\cos x}x[/math]'s. Once we’re done with that, we’ll be done with the whole problem.
[math]\frac{d}{dx}[\sin^{\cos x}x]=\sin^{\cos x}x(-\sin x\ln\sin x + \cot x\cos x) [/math]
And there it is, ladies and gentlemen! Hope you enjoyed the show!
Answer:
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I hope this will help you
