Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PRELIMINARY CALCULUS

Finddy/dxifx^3 − 3 xy+y^3 =2.

Differentiating each term in the equation with respect toxwe obtain

d

dx

(x^3 )−

d

dx

(3xy)+

d

dx

(y^3 )=

d

dx

(2),

⇒ 3 x^2 −

(

3 x

dy

dx

+3y

)

+3y^2

dy

dx

=0,

where the derivative of 3xyhas been found using the product rule. Hence, rearranging for
dy/dx,

dy

dx

=

y−x^2

y^2 −x

.

Note thatdy/dxis a function of bothxandyand cannot be expressed as a function ofx
only.

2.1.6 Logarithmic differentiation

In circumstances in which the variable with respect to which we are differentiating

is an exponent, taking logarithms and then differentiating implicitly is the simplest

way to find the derivative.

Find the derivative with respect toxofy=ax.

To find the required derivative we first take logarithms and then differentiate implicitly:

lny=lnax=xlna ⇒

1

y

dy

dx

=lna.

Now, rearranging and substituting fory, we find

dy

dx

=ylna=axlna.

2.1.7 Leibnitz’ theorem

We have discussed already how to find the derivative of a product of two or more

functions. We now considerLeibnitz’ theorem, which gives the corresponding

results for the higher derivatives of products.

Consider again the functionf(x)=u(x)v(x). We know from the product rule

thatf′=uv′+u′v. Using the rule once more for each of the products, we obtain

f′′=(uv′′+u′v′)+(u′v′+u′′v)

=uv′′+2u′v′+u′′v.

Similarly, differentiating twice more gives

f′′′=uv′′′+3u′v′′+3u′′v′+u′′′v,

f(4)=uv(4)+4u′v′′′+6u′′v′′+4u′′′v′+u(4)v.