Page:Calculus Made Easy.pdf/169

This gives

${\frac {dx}{dy}}=\epsilon ^{y}=x+a$;

hence, reverting to the original function, we get

${\frac {dy}{dx}}={\frac {1}{\;{\dfrac {dx}{dy}}\;}}={\frac {1}{x+a}}$.

 

---

 

Next try

$y=\log _{10}x$.

First change to natural logarithms by multiplying by the modulus $0.4343.$ This gives us

$y=0.4343\log _{\epsilon }x$;

whence

${\frac {dy}{dx}}={\frac {0.4343}{x}}$.

 

---

 

The next thing is not quite so simple. Try this:

$y=a^{x}$.

Taking the logarithm of both sides, we get

${\begin{aligned}\log _{\epsilon }y&=x\log _{\epsilon }a,\\{\text{ or}}\;x={\frac {\log _{\epsilon }y}{\log _{\epsilon }a}}&={\frac {1}{\log _{\epsilon }a}}\times \log _{\epsilon }y.\end{aligned}}$

Since ${\dfrac {1}{\log _{\epsilon }a}}$ is a constant, we get

${\frac {dx}{dy}}={\frac {1}{\log _{\epsilon }a}}\times {\frac {1}{y}}={\frac {1}{a^{x}\times \log _{\epsilon }a}}$;

hence, reverting to the original function.

${\frac {dy}{dx}}={\frac {1}{\;{\dfrac {dx}{dy}}\;}}=a^{x}\times \log _{\epsilon }a$.