Derivative fundamentals

Use (axⁿ)′ = anxⁿ⁻¹, the derivative of a constant is 0 and derivatives add across sums. Thus (ax² + bx + c)′ = 2ax + b. Substitute the point x for a derivative value; for f′(x) = r, solve 2ax + b = r.

The average rate of change of a function on [a, b] shows the average change in the function value when x increases by one unit. Use:

(f(b) − f(a)) / (b − a).

First find the difference between the function values, then divide it by the interval length. Pay attention to the signs of negative numbers.

The derivative f′(a) is the instantaneous rate of change and tangent slope. Tangent equation: y − f(a) = f′(a)(x − a). Find f(a) and f′(a) separately. If f′(a) = 0, the tangent is horizontal, but an extremum is not guaranteed.

Use (axⁿ)′ = anxⁿ⁻¹, the derivative of a constant is 0 and derivatives add across sums. Thus (ax² + bx + c)′ = 2ax + b. Substitute the point x for a derivative value; for f′(x) = r, solve 2ax + b = r.

The derivative f′(a) is the instantaneous rate of change and tangent slope. Tangent equation: y − f(a) = f′(a)(x − a). Find f(a) and f′(a) separately. If f′(a) = 0, the tangent is horizontal, but an extremum is not guaranteed.

Use (axⁿ)′ = anxⁿ⁻¹, the derivative of a constant is 0 and derivatives add across sums. Thus (ax² + bx + c)′ = 2ax + b. Substitute the point x for a derivative value; for f′(x) = r, solve 2ax + b = r.

Use (axⁿ)′ = anxⁿ⁻¹, the derivative of a constant is 0 and derivatives add across sums. Thus (ax² + bx + c)′ = 2ax + b. Substitute the point x for a derivative value; for f′(x) = r, solve 2ax + b = r.

Use (axⁿ)′ = anxⁿ⁻¹, the derivative of a constant is 0 and derivatives add across sums. Thus (ax² + bx + c)′ = 2ax + b. Substitute the point x for a derivative value; for f′(x) = r, solve 2ax + b = r.

Use the chain rule: ((kx + b)ⁿ)′ = n(kx + b)ⁿ⁻¹ · k. First differentiate the function, then substitute the given value of x. Do not forget the factor k obtained by differentiating the inner function.

Use the product rule: (u · v)′ = u′v + uv′. Differentiate both linear factors, form the derivative expression, and only then substitute the given value of x. Do not simply multiply the derivatives of the two factors.