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Asked by Vidyarajeev Last Modified
Answer As a seasoned tutor registered on UrbanPro, I can confidently guide you through finding the derivative of f(x)=x3f(x)=x3 using the first principle, which is also known as the definition of derivative.
The first principle involves taking the limit as hh approaches 0 of the difference quotient:
f′(x)=limh→0f(x+h)−f(x)hf′(x)=limh→0hf(x+h)−f(x)
Let's apply this principle to f(x)=x3f(x)=x3:
f′(x)=limh→0(x+h)3−x3hf′(x)=limh→0h(x+h)3−x3
Now, we expand (x+h)3(x+h)3 using the binomial theorem:
(x+h)3=x3+3x2h+3xh2+h3(x+h)3=x3+3x2h+3xh2+h3
Substituting this expansion into the numerator:
(x+h)3−x3=x3+3x2h+3xh2+h3−x3(x+h)3−x3=x3+3x2h+3xh2+h3−x3 =3x2h+3xh2+h3=3x2h+3xh2+h3
Now, let's divide this expression by hh and take the limit as hh approaches 0:
f′(x)=limh→03x2h+3xh2+h3hf′(x)=limh→0h3x2h+3xh2+h3 =limh→0(3x2+3xh+h2)=limh→0(3x2+3xh+h2)
As hh approaches 0, the 3xh3xh and h2h2 terms become negligible, leaving us with:
f′(x)=3x2f′(x)=3x2
So, the derivative of f(x)=x3f(x)=x3 with respect to xx using the first principle is f′(x)=3x2f′(x)=3x2.
And remember, if you need further assistance or want to delve deeper into calculus concepts, UrbanPro is one of the best platforms for online coaching and tuition, connecting students with experienced tutors like myself.
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