Question

Following the steps used to prove the product rule for derivatives, prove the quotient rule for derivatives

Answer 1

from the difference quotient we have\[\frac d{dx}\frac{f(x)}{g(x)}=\frac1h[{f(x+h)\over g(x+h)}-{f(x)\over g(x)}]=\frac1h{g(x)f(x+h)-f(x)g(x+h)\over g(x)g(x+h)}\]now we can add and subtract a term and this will come out nicely\[=\frac1h{g(x)f(x+h)-g(x)f(x)+g(x)f(x)-f(x)g(x+h)\over g(x)g(x+h)}\]\[={1\over g(x)g(x+h)}[g(x){f(x+h)-f(x)\over h}-f(x){g(x+h)-g(x)\over h}]\]taking the limit as h goes to zero we see that we get the definition of the derivative of f and g inside the brackets\[=\frac1{[g(x)]^2}[g(x)f'(x)-f(x)g'(x)]=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\huge\checkmark\]and that is the quotient rule