The Differentials: Example 7

Problem:Find $latex d\left[ln{\left(y^2\right)}+ln{\left(1+\sqrt y\right)}\right]&s=3&bg=ffffff&fg=000000$. SOLUTION: $latex =d\left[ln{\left(y^2\left(1+\sqrt y\right)\right)}\right]&s=1&bg=ffffff&fg=000000$$latex =d\left[ln{\left(y^2+y^\frac{5}{2}\right)}\right]&s=1&bg=ffffff&fg=000000$$latex =\frac{1}{y^2+y^\frac{5}{2}}\times\left(2y+\frac{5}{2}y^\frac{3}{2}\right)dy&s=1&bg=ffffff&fg=000000$$latex =\frac{y\left(2+\frac{5}{2}y^\frac{1}{2}\right)}{y^2\left(1+\sqrt y\right)}dy&s=1&bg=ffffff&fg=000000$$latex =\frac{2+\frac{5}{2}y^\frac{1}{2}}{y\left(1+\sqrt y\right)}dy&s=1&bg=ffffff&fg=000000$$latex =\frac{2+\frac{5}{2}\sqrt y}{y\left(1+\sqrt y\right)}dy&s=1&bg=ffffff&fg=000000$ 

The Differential

On your Calculus I, the notation $latex \frac{dy}{dx}&s=1&bg=ffffff&fg=000000$ was regarded as a single symbol to denote the derivative of a function y=f(x) with respect to x. More specifically, you have used the symbol $latex \frac{dy}{dx}&s=1&bg=ffffff&fg=000000$ to denote the limit of the quotient $latex \frac{\Delta y}{\Delta x}&s=1&bg=ffffff&fg=000000$ as $latex \Delta x&s=1&bg=ffffff&fg=000000$ approaches 0. This module introduces the concept of … Continue reading The Differential