Mathematical Methods for Physics and Engineering - Matematica.NET

PRELIMINARY CALCULUS◮Find dy/dx if x 3 − 3xy + y 3 =2.Differentiating each term in the equation with respect to x we obtainddx (x3 ) − ddx (3xy)+ ddx (y3 )= ddx (2),(⇒ 3x 2 − 3x dy )dx +3y +3y 2 dydx =0,where the derivative of 3xy has been found using the product rule. Hence, rearranging fordy/dx,dydx = y − x2y 2 − x .Note that dy/dx is a function of both x and y and cannot be expressed as a function of xonly. ◭2.1.6 Logarithmic differentiationIn circumstances in which the variable with respect to which we are differentiatingis an exponent, taking logarithms and then differentiating implicitly is the simplestway to find the derivative.◮Find the derivative with respect to x of y = a x .To find the required derivative we first take logarithms and then differentiate implicitly:ln y =lna x 1 dy= x ln a ⇒y dx =lna.Now, rearranging and substituting for y, we finddydx = y ln a = ax ln a. ◭2.1.7 Leibnitz’ theoremWe have discussed already how to find the derivative of a product of two or morefunctions. We now consider Leibnitz’ theorem, which gives the correspondingresults for the higher derivatives of products.Consider again the function f(x) =u(x)v(x). We know from the product rulethat f ′ = uv ′ + u ′ v. Using the rule once more for each of the products, we obtainf ′′ =(uv ′′ + u ′ v ′ )+(u ′ v ′ + u ′′ v)= uv ′′ +2u ′ v ′ + u ′′ v.Similarly, differentiating twice more givesf ′′′ = uv ′′′ +3u ′ v ′′ +3u ′′ v ′ + u ′′′ v,f (4) = uv (4) +4u ′ v ′′′ +6u ′′ v ′′ +4u ′′′ v ′ + u (4) v.48