Exponential, logarithmic and trigonometric derivatives
Введено| English | Значение |
|---|---|
| inner rate/ˈɪnə reɪt/ | внутренняя скорость изменения |
Which factor must appear in its derivative?
- A wave with twice the input frequency has a steeper initial slope.
- This lesson studies inner rate 内函数变化率: The derivative of the changing input inside a composite function.
The method
- For constant k and a>0: (e^(kx))′=k e^(kx), (a^(kx))′=k ln(a)a^(kx), (sin kx)′=k cos kx, (cos kx)′=−k sin kx and (tan kx)′=k sec²(kx), where sec u=1/cos u.
- Trigonometric inputs are radians.
- (ln x)′=1/x for x>0.
- Add differentiated terms and retain their coefficients.
Why does differentiating a^(kx) need the factor ln a as well as k?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Only e^(kx) differentiates to k times itself. Any other base rewrites through e, and the exponent kx ln a brings down k ln a. Forgetting ln a fails already for 2^x.
A worked case
For x>0 and cos(x/2)≠0, let f=2e^(3x)+3^(2x)−4lnx+sin2x−2cos3x+tan(x/2). Then f′=6e^(3x)+2ln3·3^(2x)−4/x+2cos2x+6sin3x+(1/2)sec²(x/2). The term −2cos3x gives a positive 6sin3x because its coefficient and derivative sign are both negative. For g=5^(−2x), g′=−2ln5·5^(−2x), so g′(0)=−2ln5. For h=sin2x−2cos3x+tan(x/2), h′(0)=2+0+1/2=5/2. For p=2e^(3x), p′(0)=6. Base a=1 gives the constant 1 and derivative zero; k=0 also produces constant functions.
Exponential, logarithmic and trigonometric derivatives
For constant k and a>0: (e^(kx))′=k e^(kx), (a^(kx))′=k ln(a)a^(kx), (sin kx)′=k cos kx, (cos kx)′=−k sin kx and (tan kx)′=k sec²(kx), where sec u=1/cos u
Choose the derivative rule and preserve every coefficient, inner rate and input restriction.
For p=2e^(3x), find p′(0).
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Retain the coefficient 2 and inner rate 3: 6e⁰=6.
Where it goes wrong
- The exponential base is constant but its exponent changes: a^(kx) needs ln a as well as k.
- The derivative of ln x is 1/x, not ln x/x.
- Tangent is undefined where cos(kx)=0 and its derivative retains those exclusions.
- Derivative formulas in radians need π/180 factors for degree variables.
- A negative k changes the inner-rate sign; it must not disappear.
Not true: The derivative of a^(kx) is k a^(kx) for every positive base a. The derivative of a^(kx) is k·ln a·a^(kx); the factor ln a is 1 only for base e.
For the worked h, find h′(0).
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
2cos0+6sin0+(1/2)sec²0=5/2.
The derivative of a^(kx) is k·ln a·a^(kx).
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. The derivative of a^(kx) is k·ln a·a^(kx); the factor ln a is 1 only for base e.
Applying it
- Identify the function family, preserve its coefficient, differentiate the outer function and multiply by k.
- List log and tangent domain exclusions before evaluating at a point.
- Check the sign against an increasing/declining exponential or the local wave direction.
- Derivatives of linear combinations are obtained by addition, not by multiplying terms.
For y=ln x, find its derivative at x=4.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The positive input 4 gives 1/4.
Match each function to its derivative.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Trigonometric inputs are in radians; every composite term keeps its coefficient and its inner-rate factor, including sign.
In your course
- AQA A-level Mathematics (7357), specification section G.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Inner rate: The derivative of the changing input inside a composite function.
- For constant k and a>0: (e^(kx))′=k e^(kx), (a^(kx))′=k ln(a)a^(kx), (sin kx)′=k cos kx, (cos kx)′=−k sin kx and (tan kx)′=k sec²(kx), where sec u=1/cos u.
- Watch for: the exponential base is constant but its exponent changes: a^(kx) needs ln a as well as k.