Derivatives of Exponential, Logarithmic and Trigonometric Functions

For y = a^x with a positive constant base, the ratio of gradient to height is ln a. The height is a^x and the gradient is a^x ln a. With base e, ln e = 1, so the gradient equals the height everywhere.

This uniqueness concerns the base of a^x. It does not mean e^x is the only function equal to its derivative. Every constant multiple C e^x also has derivative C e^x.

To differentiate an exponential composition, retain the full exponent and multiply by its derivative. For e^(2x+1), this gives 2e^(2x+1).

For logarithms, begin with the domain. The function ln(g(x)) is real only where g(x) > 0. On that domain its derivative is g'(x)/g(x). For ln(8x+3), the domain is x > -3/8 and the derivative is 8/(8x+3). Simplifying a derivative does not add inputs to the original function's domain.

In radians, sine has derivative cosine, and cosine has derivative negative sine. A derivative gives the slope of the tangent at each input.

To find a slope at one point, differentiate before evaluating. For y = cos x at x = pi, the derivative is -sin x, so the slope is -sin pi = 0. The tangent is horizontal there.

An inner function contributes another factor. For sin(6x), differentiating sine gives cos(6x), and differentiating the argument gives 6. Their product is 6 cos(6x). Changing the argument to 6 would describe a different function.

At the origin, the sine curve passes through (0, 0) with slope cos 0 = 1. A line with that point and slope is y = x. It gives the local estimate sin x ≈ x. The equality is approximate away from zero, and x must be measured in radians.

Look at the final operation used to build a function. Multiplying two functions calls for the product rule. Applying one function to another calls for the chain rule. One expression can need both.

For x ln(x+2), the real domain is x > -2. Treat x and ln(x+2) as the two product factors. Their derivatives are 1 and 1/(x+2). The logarithmic factor uses the chain rule, with inner derivative 1.

The product rule gives 1 times ln(x+2) plus x times 1/(x+2). The result is ln(x+2) + x/(x+2). Multiplying the two derivatives would omit both required product-rule terms.

For a composition, keep the original inner argument inside the outer derivative. Then multiply by the inner derivative. For a logarithm, also check that the argument is positive. For trigonometric rules in this topic, use radians.

Establish the limits using unit-circle geometry. Let O = (0,0), A = (1,0), B = (cos h,sin h) and T = (1,tan h), where 0 < h < pi/2 radians. Triangle OAB lies inside the sector between OA and OB, which lies inside triangle OAT. Their areas are (sin h)/2, h/2 and (tan h)/2. Therefore sin h <= h <= tan h, giving cos h <= sin h/h <= 1.

The sine bound and odd symmetry give |sin u| <= |u| near zero. The identity 1-cos h = 2sin²(h/2) then gives 0 <= 1-cos h <= h²/2. Hence cos h → 1. The squeeze theorem gives sin h/h → 1 from the right. The quotient sin h/h is even, so its left-hand limit is also 1. Also |(cos h-1)/h| <= |h|/2, so this quotient tends to 0. Knowing only cos h → 1 would not establish that quotient limit.

For fixed x, the identity sin(x+h) = sin x cos h + cos x sin h rewrites [sin(x+h)-sin x]/h as sin x[(cos h-1)/h] + cos x[sin h/h]. Its limit is sin x times 0 plus cos x times 1. Thus (sin x)' = cos x.

Similarly, cos(x+h) = cos x cos h - sin x sin h rewrites the cosine difference quotient as cos x[(cos h-1)/h] - sin x[sin h/h]. Its limit is -sin x.

After proving these rules, sin 0 = 0 and cos 0 = 1 give the sine tangent y = x and the estimate sin x ≈ x near zero. The geometric argument establishes the limits without using that tangent or a Taylor series.