What is the derivative of this function?
$$ f(x)= \cos \bigg(\frac{\sin x}{x}\bigg) $$
What is the derivative of this function?
$$ f(x)= \cos \bigg(\frac{\sin x}{x}\bigg) $$
I’m assuming that you mean that $$f(x)=\cos\left(\frac{\sin x}x\right)\;.$$ Just apply the chain rule. At the outermost level this is a cosine, so differentiate it as such:
$$f\,'(x)=-\sin\left(\frac{\sin x}x\right)\cdot\left(\frac{\sin x}x\right)'\;,$$
where the prime denotes the derivative with respect to $x$. Now you need $\left(\frac{\sin x}x\right)'$; the function is a quotient, so use the quotient rule:
$$\left(\frac{\sin x}x\right)'=\frac{x(\sin x)'-(\sin x)x'}{x^2}=\frac{x\cos x-\sin x}{x^2}\;.$$
Now just put the pieces together.
You work from the outside in, slowly apply the chain rule. First, what is the derivative of $\cos$? Well that's $-\sin x$. Thus far we have
$$f'(x)=-sin\bigg(\frac{\sin x}{x}\bigg) \bigg(\frac{\sin x}{x}\bigg)'$$
where $\bigg(\frac{\sin x}{x} \bigg)'$ is the derivative of the 'stuff' on the inside (this came from the chain rule). So we need the derivative of $\frac{\sin x}{x}$. Use the quotient rule. You will get $$\frac{x \cos x-\sin x}{x^2}$$ Now we multiply these to get the answer $$f'(x)=-\sin\bigg(\frac{\sin x}{x} \bigg)\frac{x \cos x-\sin x}{x^2}$$
HINT Use chain rule to differentiate. Remember the general "formula" for the chain rule is: $$f' = g'(h(x)) * h'(x)$$
In this case, your $g(x)$ is $\cos x$ and your $h(x)$ is $\frac{\sin x}{x}$. You will also have to use to quotient rule to differentiate your $\frac{\sin x}{x}$ term. Recall that quotient rule is:
$$\frac{f'g - fg'}{g^2}$$
So in effect, you have
$$f' = g'(h(x)) * \frac{f'g - fg'}{g^2} $$
HINT 2 The derivative of $\cos x$ is $-\sin x$ and the derivative of $\sin x$ is $\cos x$
Can you take it from here?