Derivative Calculator

Derivative of $x\,\cot\left(x\right)$

Differentiated with respect to x. Edit the function · Differentiation rules

Step-by-step solution

Input function
$$x\,\cot\left(x\right)$$
Start of the differentiation
$$\frac{d}{dx}{\left[x\,\cot\left(x\right)\right]}$$

Yellow: the expression being replaced. Green: its replacement.

Step 1 — Product Rule Differentiation rule
📖 Rule
$$\frac{d}{dx}(u \cdot v)=u'v+uv'$$
With:
  • $u = x$
  • $v = \cot\left(x\right)$

Why: First, differentiate the first factor and multiply by the unchanged second factor. Add the unchanged first factor times the derivative of the second. Do not simply multiply the two derivatives.

How does this rule work? Wikipedia
🧮 Current expression
$$\bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left[x\,\cot\left(x\right)\right]}\,}}}$$
✨ After: Product Rule
$$\bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left[x\right]}\,\cot\left(x\right) + x\,\frac{d}{dx}{\left[\cot\left(x\right)\right]}\,}}}$$
Step 2 — Derivative of a Variable Differentiation rule
📖 Rule
$$\frac{d}{dx}(x)=1$$

Why: When x increases by one, the value of this term also increases by one. Its derivative is therefore one.

Changed part: $\frac{d}{dx}{\left[x\right]} \;\Longrightarrow\; 1$
How does this rule work?
🧮 Current expression
$$\bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left[x\right]}\,}}}\,\cot\left(x\right) + x\,\frac{d}{dx}{\left[\cot\left(x\right)\right]}$$
✨ After: Derivative of a Variable
$$\bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,1\,}}}\,\cot\left(x\right) + x\,\frac{d}{dx}{\left[\cot\left(x\right)\right]}$$
Step 3 — Remove a factor of one Simplification
📖 Rule
$$1\cdot a=a,\quad \frac{a}{1}=a$$

Why: Multiplication or division by one leaves the value unchanged.

Changed part: $1\,\cot\left(x\right) \;\Longrightarrow\; \cot\left(x\right)$
🧮 Current expression
$$\bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,1\,\cot\left(x\right)\,}}} + x\,\frac{d}{dx}{\left[\cot\left(x\right)\right]}$$
✨ After: Remove a factor of one
$$\bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\cot\left(x\right)\,}}} + x\,\frac{d}{dx}{\left[\cot\left(x\right)\right]}$$
Step 4 — Derivative of the cotangent Differentiation rule
📖 Rule
$$\frac{d}{dx}\cot\left(x\right)=-\frac{1}{\sin\left(x\right)^{2}}$$

Why: The argument is exactly x. Apply the known elementary derivative.

Changed part: $\frac{d}{dx}{\left[\cot\left(x\right)\right]} \;\Longrightarrow\; -\frac{1}{\sin\left(x\right)^{2}}$
How does this rule work? Wikipedia
🧮 Current expression
$$\cot\left(x\right) + x\,\bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left[\cot\left(x\right)\right]}\,}}}$$
✨ After: Derivative of the cotangent
$$\cot\left(x\right) + x\,\bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\left(-\frac{1}{\sin\left(x\right)^{2}}\right)\,}}}$$
Step 5 — Determine the sign of the product Simplification
📖 Rule
$$a\cdot(-b)=-(a\cdot b),\quad (-a)\cdot(-b)=a\cdot b$$

Why: Move the minus signs of the factors in front of the product. An even number of minus signs gives plus, an odd number gives minus.

Changed part: $x\,\left(-\frac{1}{\sin\left(x\right)^{2}}\right) \;\Longrightarrow\; -x \cdot \frac{1}{\sin\left(x\right)^{2}}$
🧮 Current expression
$$\cot\left(x\right) + \bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,x\,\left(-\frac{1}{\sin\left(x\right)^{2}}\right)\,}}}$$
✨ After: Determine the sign of the product
$$\cot\left(x\right) + \bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\left(-x \cdot \frac{1}{\sin\left(x\right)^{2}}\right)\,}}}$$
Step 6 — Combine into one fraction Simplification
📖 Rule
$$a\cdot\frac{b}{c}=\frac{a\cdot b}{c}$$

Why: A factor in front of a fraction multiplies its numerator, giving a single fraction.

Changed part: $-x \cdot \frac{1}{\sin\left(x\right)^{2}} \;\Longrightarrow\; -\frac{x}{\sin\left(x\right)^{2}}$
🧮 Current expression
$$\cot\left(x\right) + \bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\left(-x \cdot \frac{1}{\sin\left(x\right)^{2}}\right)\,}}}$$
✨ After: Combine into one fraction
$$\cot\left(x\right) + \bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\left(-\frac{x}{\sin\left(x\right)^{2}}\right)\,}}}$$
Step 7 — Combine plus and minus Simplification
📖 Rule
$$a+(-b)=a-b$$

Why: Adding a negative expression is the same as subtracting the positive expression.

🧮 Current expression
$$\bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\cot\left(x\right) + \left(-\frac{x}{\sin\left(x\right)^{2}}\right)\,}}}$$
✨ After: Combine plus and minus
$$\bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\cot\left(x\right) - \frac{x}{\sin\left(x\right)^{2}}\,}}}$$
Final result
$$\cot\left(x\right) - \frac{x}{\sin\left(x\right)^{2}}$$
Valid where:
  • $\sin\left(x\right)\ne0$ – The function and its derivative are undefined at poles.

The derivative is only valid where the original function is defined.

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Cross-check with a computer algebra system

The derivative computed directly by the computer algebra system Maxima:

$$\cot\left(x\right) - x\,\csc\left(x\right)^{2}$$
Step-by-step result ↔ Maxima: identical ✓