How do you find f'(x) using the definition of a derivative for f(x)=(1-6t)/(5+t)f(x)=1−6t5+t?

1 Answer
Dharma R.
Sep 23, 2015

-31/(5+t)^2−31(5+t)2

Explanation:

f(t)=(1-6t)/(5+t)f(t)=1−6t5+t
f(t+h)=(1+6(t+h))/(5+(t+h))f(t+h)=1+6(t+h)5+(t+h)
f(t+h)-f(t)=(1-6(t+h))/(5+(t+h))-(1-6t)/(5+t)=(5+t-6(t^2+5h+5t+ht)-5-t-h+30t+6t^2+6ht)/((5+t+h)(5+t))=(-31h)/((5+t+h)(5+t))f(t+h)−f(t)=1−6(t+h)5+(t+h)−1−6t5+t=5+t−6(t2+5h+5t+ht)−5−t−h+30t+6t2+6ht(5+t+h)(5+t)=−31h(5+t+h)(5+t)
Lt_(h->0)((f(t+h)-f(h))/h)=Lt_(h->0)(-31h)/(h(5+t+h)(5+t))=Lt_(h->0)(-31)/((5+t+h)(5+t))=-31/(5+t)^2Lth→0(f(t+h)−f(h)h)=Lth→0−31hh(5+t+h)(5+t)=Lth→0−31(5+t+h)(5+t)=−31(5+t)2

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