Applications of the Theorem for Implicit Differentiation

Excerpts

• Applications of the Theorem for Implicit Differentiation

  We must leave $\dfrac{da}{dx}$ and $\dfrac{db}{dx}$ as we find them, until we know *what* functions $a$ and $b$ are of $x$; but as we know what function $z$ is of $a$ and $b$, we substitute for $\dfrac{dz}{da}$ and $\dfrac{dz}{db}$.

• Applications of the Theorem for Implicit Differentiation

  Let $z = \dfrac{a}{b}$. If $a$ become $a + da$, $z$ becomes $\dfrac{a + da}{b}$ or $\dfrac{a}{b} + \dfrac{da}{b}$, and $\dfrac{dz}{da}$ is $\dfrac{1}{b}$.

• Applications of the Theorem for Implicit Differentiation

  Again, $a^{b+db} = a^{b}\, a^{db} = a^{b}(1 + \log a\, db + \etc.)$ whence $\dfrac{dz}{db} = a^{b} \log a$.

• Applications of the Theorem for Implicit Differentiation

  In this case, and part of the following, the limiting ratio of the increments is the same as that of the increments themselves.

Equations

• Applications of the Theorem for Implicit Differentiation

  z = ab

  Defines z as the product of the two functions a and b of x.

• Applications of the Theorem for Implicit Differentiation

  \frac{dz}{dx} = \frac{dz}{da}\, \frac{da}{dx} + \frac{dz}{db}\, \frac{db}{dx}

  The derivative of z with respect to x, where z depends on x indirectly through a and b, is the sum of the partial rates through each intermediate function.

• Applications of the Theorem for Implicit Differentiation

  \dfrac{dz}{db} = b

  For z = ab, the ratio dz/db equals b. The book writes this as dz/db = b, but the argument given (z becomes ab + b da when a becomes a + da) yields dz/da = b; the subscript appears to be an erratum, flagged here and not corrected.

• Applications of the Theorem for Implicit Differentiation

  \dfrac{dz}{db} = a

  For z = ab, the ratio dz/db equals a (stated as similar to the previous case).

• Applications of the Theorem for Implicit Differentiation

  \frac{dz}{dx} = b\, \frac{da}{dx} + a\, \frac{db}{dx}

  The derivative of a product ab with respect to x is b times da/dx plus a times db/dx.

• Applications of the Theorem for Implicit Differentiation

  z = \dfrac{a}{b}

  Defines z as the quotient of a and b.

• Applications of the Theorem for Implicit Differentiation

  \frac{dz}{dx} = \frac{1}{b}\, \frac{da}{dx} - \frac{a}{b^{2}}\, \frac{db}{dx}

  The derivative of the quotient a/b with respect to x, with the partial rates dz/da = 1/b and dz/db = -a/b^2 substituted into the general formula.

• Applications of the Theorem for Implicit Differentiation

  z = a^{b}

  Defines z as a raised to the power b, where both a and b may be functions of x.

• Applications of the Theorem for Implicit Differentiation

  (a + da)^{b} = a^{b} + ba^{b-1}\, da + \etc.

  Binomial expansion to first order of (a + da)^b, giving the increment of a^b when a is increased by da.

• Applications of the Theorem for Implicit Differentiation

  \dfrac{dz}{da} = ba^{b-1}

  For z = a^b, the ratio dz/da equals b times a to the power b minus 1.

• Applications of the Theorem for Implicit Differentiation

  a^{b+db} = a^{b}\, a^{db} = a^{b}(1 + \log a\, db + \etc.)

  Expansion of a^(b+db) as a^b times a^db, with a^db approximated to first order by 1 + log a times db.

• Applications of the Theorem for Implicit Differentiation

  \dfrac{dz}{db} = a^{b} \log a

  For z = a^b, the ratio dz/db equals a^b times the logarithm of a.

• Applications of the Theorem for Implicit Differentiation

  \frac{dz}{dx} = ba^{b-1}\, \frac{da}{dx} + a^{b} \log a\, \frac{db}{dx}

  The derivative of a^b with respect to x when both a and b are functions of x.

Problems

No exercises in this chapter.