Summary
Highlights
The video starts by explaining the derivative of a reciprocal, given by the formula dy/dX = - (dV/dX) / V^2, where V is the function in the denominator. An example demonstrates calculating the derivative of y = 1/(x+1), resulting in -1/(x+1)^2. Another example covers y = 1/(3t^2 + 4t - 5), yielding -(6t+4)/(3t^2 + 4t - 5)^2.
This section introduces the derivative of functions with rational exponents, using the formula dy/dX = (1/m) * X^(1/m - 1), where m is not equal to zero. The first example is y = sqrt(x) + cube root(x), which is rewritten as X^(1/2) + X^(1/3). The derivative is calculated as (1/2)X^(-1/2) + (1/3)X^(-2/3), then simplified to 1/(2*sqrt(x)) + 1/(3*cube root(x^2)). The second example is y = sqrt(x) + cube root(x^2), transformed to X^(1/2) + X^(2/3). The derivative is (1/2)X^(-1/2) + (2/3)X^(-1/3), simplified to 1/(2*sqrt(x)) + 2/(3*cube root(x)).
The chain rule is introduced with the formula dy/dX = (dy/dU) * (dU/dX). The first example is y = (4x^2 - 2x + 5)^3. By setting U = 4x^2 - 2x + 5, then Y = U^3. The derivatives dY/dU = 3U^2 and dU/dX = 8x - 2 are found. Substituting back, the final derivative is 3(4x^2 - 2x + 5)^2 * (8x - 2).
A second example for the chain rule is presented: y = cube root((t^2 - 2t + 3)^3), which is converted to Y = U^(3/2) where U = t^2 - 2t + 3. The derivatives dY/dU = (3/2)U^(1/2) and dU/dT = 2t - 2 are found. The chain rule is applied for dY/dT, leading to (3/2)(t^2 - 2t + 3)^(1/2) * (2t - 2). This is further simplified to 3(t - 1) * sqrt(t^2 - 2t + 3).