Question

differentiate.
y = 2x² cos x cot x
y =

Explanation:

Step1: Recall Product Rule

The product rule for differentiation is \((uvw)' = u'vw + uv'w + uvw'\), where \(u = 2x^2\), \(v=\cos x\), \(w = \cot x\).

Step2: Differentiate \(u = 2x^2\)

Using the power rule \((x^n)'=nx^{n - 1}\), we get \(u'=4x\).

Step3: Differentiate \(v=\cos x\)

The derivative of \(\cos x\) is \(-\sin x\), so \(v'=-\sin x\).

Step4: Differentiate \(w = \cot x\)

The derivative of \(\cot x\) is \(-\csc^{2}x\), so \(w'=-\csc^{2}x\).

Step5: Apply Product Rule

\[

$$\begin{align*} y'&=u'vw+uv'w+uvw'\\ &=4x\cdot\cos x\cdot\cot x+2x^{2}\cdot(-\sin x)\cdot\cot x+2x^{2}\cdot\cos x\cdot(-\csc^{2}x) \end{align*}$$

\]
Simplify each term:

• First term: \(4x\cos x\cot x\)
• Second term: \(2x^{2}(-\sin x)\cot x=- 2x^{2}\sin x\cdot\frac{\cos x}{\sin x}=-2x^{2}\cos x\) (since \(\cot x=\frac{\cos x}{\sin x}\))
• Third term: \(2x^{2}\cos x(-\csc^{2}x)=-2x^{2}\cos x\cdot\frac{1}{\sin^{2}x}\) (since \(\csc x = \frac{1}{\sin x}\))

Now, rewrite the first term using \(\cot x=\frac{\cos x}{\sin x}\): \(4x\cos x\cdot\frac{\cos x}{\sin x}=\frac{4x\cos^{2}x}{\sin x}\)

The second term is \(-2x^{2}\cos x\) and the third term is \(-\frac{2x^{2}\cos x}{\sin^{2}x}\)

We can also factor out common terms or rewrite in terms of sine and cosine:

\[
y'=4x\cos x\cot x-2x^{2}\cos x - 2x^{2}\cos x\csc^{2}x
\]
Or further simplify by combining terms over a common denominator (though the above form is also correct). Another way is to use trigonometric identities:

Since \(\cot x=\frac{\cos x}{\sin x}\), the original function \(y = 2x^{2}\cos x\cdot\frac{\cos x}{\sin x}=\frac{2x^{2}\cos^{2}x}{\sin x}\)

Now differentiate \(y=\frac{2x^{2}\cos^{2}x}{\sin x}\) using quotient rule \((\frac{f}{g})'=\frac{f'g - fg'}{g^{2}}\), where \(f = 2x^{2}\cos^{2}x\) and \(g=\sin x\)

First, find \(f'\): Using product rule on \(f = 2x^{2}\cos^{2}x\), let \(a = 2x^{2}\), \(b=\cos^{2}x\)

\(a' = 4x\), \(b'=2\cos x(-\sin x)=-2\cos x\sin x\)

So \(f'=4x\cos^{2}x+2x^{2}(-2\cos x\sin x)=4x\cos^{2}x - 4x^{2}\cos x\sin x\)

\(g'=\cos x\)

Then by quotient rule:
\[

$$\begin{align*} y'&=\frac{(4x\cos^{2}x - 4x^{2}\cos x\sin x)\sin x-2x^{2}\cos^{2}x\cdot\cos x}{\sin^{2}x}\\ &=\frac{4x\cos^{2}x\sin x-4x^{2}\cos x\sin^{2}x - 2x^{2}\cos^{3}x}{\sin^{2}x}\\ &=\frac{4x\cos^{2}x\sin x-2x^{2}\cos x(2\sin^{2}x+\cos^{2}x)}{\sin^{2}x} \end{align*}$$

\]
But the form from the product rule is also acceptable. The simplified form using product rule and trigonometric identities:

\[
y' = 4x\cos x\cot x-2x^{2}\cos x-2x^{2}\cos x\csc^{2}x
\]
Or we can write all terms with sine in the denominator:

\[
y'=\frac{4x\cos^{2}x}{\sin x}-2x^{2}\cos x-\frac{2x^{2}\cos x}{\sin^{2}x}
\]

Answer:

\(4x\cos x\cot x - 2x^{2}\cos x-2x^{2}\cos x\csc^{2}x\) (or equivalent simplified forms)