How Do You Find the Derivative of This Function?

[riri]

Hello, I am having trouble finding the derivative of this function. Any help would be appreciated!

\(\displaystyle \frac{x\cos(x)+\sqrt{x}}{x+\sin(x)}\)

What I tried was expanding it first to like \(\displaystyle (x+\sin(x)(x\cos(x)+\sqrt{x})'-(x\cos(x)+\sqrt{x}\)
But I ended up with a long weird answer and doesn't seem to work!

Can anyone give ideas for this?

 

[MarkFL]

Hello, riri! :D

I have moved this thread here to our Calculus forum. While the problem does involve trigonometric functions, it is primarily a problem in differential calculus.

Also, I have edited your post to wrap your $\LaTeX$ code in [MATH][/MATH] tags.

We are given:

\(\displaystyle f(x)=\frac{x\cos(x)+\sqrt{x}}{x+\sin(x)}\)

We have a quotient, so I would suggest applying the quotient rule to get:

\(\displaystyle f'(x)=\frac{\left(x+\sin(x)\right)\left(x\cos(x)+\sqrt{x}\right)^{\prime}-\left(x\cos(x)+\sqrt{x}\right)\left(x+\sin(x)\right)^{\prime}}{\left(x+\sin(x)\right)^2}\)

Can you continue?

 

[riri]

Hi Mark! :D
Thank you, as you can see I'm still pretty new to this website but love it so much!

Yup, so I did use the quotient rule and got up to that part, it's just that I'm stuck on what to do after like:

\(\displaystyle \frac{x+\sin(x)(1(-\sin(x))+{x}^{\frac{-1}{2}}-(x\cos(x)+\sqrt{x}(1+\cos(x)}{x^2+\sin(x)^2}\)

I feel the direction I'm going in is wrong because I got a really long equation thing and it's stuck there... I'm hoping there's a simple way to solve this though!

Thank you!

 

[MarkFL]

Let's first look at the numerator:

\(\displaystyle \left(x+\sin(x)\right)\left(x\cos(x)+\sqrt{x}\right)^{\prime}-\left(x\cos(x)+\sqrt{x}\right)\left(x+\sin(x)\right)^{\prime}\)

Now, let's look at the parts still left to differentiate, beginning with:

\(\displaystyle \left(x\cos(x)+\sqrt{x}\right)^{\prime}\)

For the first term, we will need to use the product rule:

\(\displaystyle \frac{d}{dx}\left(x\cos(x)\right)=x(-\sin(x))+1(\cos(x))=\cos(x)-x\sin(x)\)

And for the second term, we need the power rule:

\(\displaystyle \frac{d}{dx}\left(\sqrt{x}\right)=\frac{d}{dx}\left(x^{\frac{1}{2}}\right)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\)

Putting this together, we may write:

\(\displaystyle \left(x\cos(x)+\sqrt{x}\right)^{\prime}=\cos(x)-x\sin(x)+\frac{1}{2\sqrt{x}}\)

The other part still left to differentiate is:

\(\displaystyle \left(x+\sin(x)\right)^{\prime}=\frac{d}{dx}\left(x+\sin(x)\right)=1+\cos(x)\)

And so the numerator then becomes:

\(\displaystyle \left(x+\sin(x)\right)\left(\cos(x)-x\sin(x)+\frac{1}{2\sqrt{x}}\right)-\left(x\cos(x)+\sqrt{x}\right)\left(1+\cos(x)\right)\)

You could multiply everything out and see if you can combine terms and use trig. identities to simplify.

For the denominator, you have committed an error so commonly made, it has a special name, which is "The Freshman's Dream." You have stated essentially:

\(\displaystyle (a+b)^2=a^2+b^2\)

This is not true in general, what you want, if you are going to expand it, is:

\(\displaystyle (a+b)^2=a^2+2ab+b^2\)