Proof
- Take logs: log y = x log x.
- Differentiate: (1/y) dy/dx = log x + x × (1/x) = log x + 1.
- So dy/dx = xˣ(1 + log x).
Standard derivatives
| Function | Derivative |
|---|---|
| d/dx(sin x) | cos x |
| d/dx(cos x) | −sin x |
| d/dx(tan x) | sec²x |
| d/dx(cot x) | −cosec²x |
| d/dx(sec x) | sec x tan x |
| d/dx(cosec x) | −cosec x cot x |
| d/dx(eˣ) | eˣ |
| d/dx(aˣ) | aˣ log a |
| d/dx(log x) | 1/x |
| d/dx(xⁿ) | n xⁿ⁻¹ |
| d/dx(√x) | 1/(2√x) |
| d/dx(1/x) | −1/x² |
| d/dx(sin⁻¹x) | 1/√(1 − x²) |
| d/dx(cos⁻¹x) | −1/√(1 − x²) |
| d/dx(tan⁻¹x) | 1/(1 + x²) |
| d/dx(cot⁻¹x) | −1/(1 + x²) |
| d/dx(sec⁻¹x) | 1/(|x|√(x² − 1)) |
| d/dx(xˣ) | xˣ(1 + log x) |
| d/dx(logₐx) | 1/(x log a) |
| d/dx(sin²x) | sin 2x |
| d/dx(log(sin x)) | cot x |
| d/dx(|x|) | x/|x| (x ≠ 0) |
Questions students ask
What is the derivative of xˣ?
d/dx(xˣ) = xˣ(1 + log x).
How do you differentiate xˣ?
Take logs: log y = x log x. Differentiate: (1/y) dy/dx = log x + x × (1/x) = log x + 1. So dy/dx = xˣ(1 + log x).