Derivative Calculator

Derivative of $\mathrm{e}^{x}\,\cosh\left(x\right)$

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

Step-by-step solution

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

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:
  • $u = \mathrm{e}^{x}$
  • $v = \cosh\left(x\right)$

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
$$\bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left[\mathrm{e}^{x}\,\cosh\left(x\right)\right]}\,}}}$$
✨ After: Product Rule
$$\bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left[\mathrm{e}^{x}\right]}\,\cosh\left(x\right) + \mathrm{e}^{x}\,\frac{d}{dx}{\left[\cosh\left(x\right)\right]}\,}}}$$
Step 2 — Exponential Rule Differentiation rule
📖 Rule
$$\frac{d}{dx}(e^{u})=e^{u}\cdot u'$$
With:
  • $u = x$

Why: The exponential with base e remains unchanged. For a composite exponent, multiply by its derivative using the chain rule.

Changed part: $\frac{d}{dx}{\left[\mathrm{e}^{x}\right]} \;\Longrightarrow\; \mathrm{e}^{x}\,\frac{d}{dx}{\left[x\right]}$
How does this rule work? Wikipedia
🧮 Current expression
$$\bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left[\mathrm{e}^{x}\right]}\,}}}\,\cosh\left(x\right) + \mathrm{e}^{x}\,\frac{d}{dx}{\left[\cosh\left(x\right)\right]}$$
✨ After: Exponential Rule
$$\bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\mathrm{e}^{x}\,\frac{d}{dx}{\left[x\right]}\,}}}\,\cosh\left(x\right) + \mathrm{e}^{x}\,\frac{d}{dx}{\left[\cosh\left(x\right)\right]}$$
Step 3 — 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
$$\mathrm{e}^{x}\,\bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left[x\right]}\,}}}\,\cosh\left(x\right) + \mathrm{e}^{x}\,\frac{d}{dx}{\left[\cosh\left(x\right)\right]}$$
✨ After: Derivative of a Variable
$$\mathrm{e}^{x} \cdot \bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,1\,}}}\,\cosh\left(x\right) + \mathrm{e}^{x}\,\frac{d}{dx}{\left[\cosh\left(x\right)\right]}$$
Step 4 — 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: $\mathrm{e}^{x} \cdot 1 \;\Longrightarrow\; \mathrm{e}^{x}$
🧮 Current expression
$$\bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\mathrm{e}^{x} \cdot 1\,}}}\,\cosh\left(x\right) + \mathrm{e}^{x}\,\frac{d}{dx}{\left[\cosh\left(x\right)\right]}$$
✨ After: Remove a factor of one
$$\bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\mathrm{e}^{x}\,}}}\,\cosh\left(x\right) + \mathrm{e}^{x}\,\frac{d}{dx}{\left[\cosh\left(x\right)\right]}$$
Step 5 — Derivative of the hyperbolic cosine Differentiation rule
📖 Rule
$$\frac{d}{dx}\cosh\left(x\right)=\sinh\left(x\right)$$

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

Changed part: $\frac{d}{dx}{\left[\cosh\left(x\right)\right]} \;\Longrightarrow\; \sinh\left(x\right)$
How does this rule work? Wikipedia
🧮 Current expression
$$\mathrm{e}^{x}\,\cosh\left(x\right) + \mathrm{e}^{x}\,\bbox[#fff1b8, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\frac{d}{dx}{\left[\cosh\left(x\right)\right]}\,}}}$$
✨ After: Derivative of the hyperbolic cosine
$$\mathrm{e}^{x}\,\cosh\left(x\right) + \mathrm{e}^{x}\,\bbox[#d6f5df, padding:0.15em]{{\color{black}{\vphantom{\dfrac{d}{dx}}\,\sinh\left(x\right)\,}}}$$
Final result
$$\mathrm{e}^{x}\,\cosh\left(x\right) + \mathrm{e}^{x}\,\sinh\left(x\right)$$
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Cross-check with a computer algebra system

The derivative computed directly by the computer algebra system Maxima:

$$\mathrm{e}^{x}\,\sinh\left(x\right) + \mathrm{e}^{x}\,\cosh\left(x\right)$$
Step-by-step result ↔ Maxima: identical ✓