Question Video: Calculating the Derivative at a Point | Nagwa Question Video: Calculating the Derivative at a Point | Nagwa

Question Video: Calculating the Derivative at a Point Mathematics

Find the derivative of 𝑓(𝑥) = 𝑥² at the point 𝑥 = 2 from first principles.

02:16

Video Transcript

Find the derivative of the function 𝑓 of 𝑥 is equal to 𝑥 squared at the point 𝑥 is equal to two from first principles.

The question wants us to find the derivative of the function 𝑓 of 𝑥 is equal to 𝑥 squared when 𝑥 is equal to two from first principles. We’ll start by recalling what we mean by the derivative of a function 𝑓 of 𝑥 at the point 𝑥 naught. This is given by 𝑓 prime of 𝑥 naught is equal to the limit as ℎ approaches zero of 𝑓 evaluated at 𝑥 naught plus ℎ minus 𝑓 evaluated at 𝑥 naught divided by ℎ as long as this limit exists. From the question, we see we want to find the derivative of the function 𝑓 of 𝑥 is equal to 𝑥 squared when 𝑥 is equal to two. So we’ll set 𝑥 naught equal to two.

Substituting 𝑥 naught is equal to two into our definition of the derivative gives us 𝑓 prime of two is equal to the limit as ℎ approaches zero of 𝑓 evaluated at two plus ℎ minus 𝑓 evaluated at two divided by ℎ provided this limit exists. Now, remember our function 𝑓 of 𝑥 is 𝑥 squared. So 𝑓 evaluated at two plus ℎ is two plus ℎ all squared and 𝑓 evaluated at two is two squared. So we now need to evaluate the limit as ℎ approaches zero of two plus ℎ squared minus two squared divided by ℎ.

To evaluate this limit, we’ll start by distributing the square over our parentheses. We have two plus ℎ all squared is equal to four plus four ℎ plus ℎ squared. So now we want to evaluate the limit as ℎ approaches zero of four plus ℎ plus ℎ squared minus four divided by ℎ. We can simplify the numerator in this limit since four minus four is equal to zero. So now we just want the limit as ℎ approaches zero of four ℎ plus ℎ squared divided by ℎ. And since our limit is as ℎ is approaching zero, ℎ is not equal to zero. So we can simplify this limit by canceling the shared factors of ℎ on our numerator and our denominator.

This gives us the limit as ℎ approaches zero of four plus ℎ. And this is a linear function, so we can evaluate this by direct substitution. Substituting ℎ is equal to zero, we get four plus zero, which is of course just equal to four. Therefore, we’ve shown from first principles if 𝑓 of 𝑥 is equal to 𝑥 squared, then 𝑓 prime of two is equal to four.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy