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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 11: Basic differentiation rules

# Proofs of the constant multiple and sum/difference derivative rules

Constant multiple rule says that the derivative of f(x)=kg(x) is f'(x)=kg'(x). Sum/Difference rule says that the derivative of f(x)=g(x)±h(x) is f'(x)=g'(x)±h'(x). Sal introduces and justifies these rules.

## Want to join the conversation?

• @ where did you get the h from?
• We get that h from the definition of a derivative:
f'(x) = lim(h->0) ( (f(x+h)-f(x)) / (x+h)-x)
'h' basically represents a very little increase in x value.
instead of 'h' some book also have (delta x) [delta looks like a small triangle, its a greek symbol]
• At Sal chooses algebraic argument over graphical one, but in the end boils it down to limits' properties, which weren't proved here, and given as "common sense", which is okay with me.. but maybe there is a more thorough proof that includes limits' properties?
• Yes, although a more typical notation would be something like `dx1/dt - dx2/dt = d/dt [ x1 - x2 ]`.

We often use the d/dt (or d/dx, etc.) as a kind of operator to indicate that we're taking the derivative of whatever function comes after the d/dt.

Hope that helps.
• random question over rigorous-ness in mathematics. How do we know that 1+1=2?
(1 vote)
• We know it from an area of mathematics known as Foundation of Mathematics. This is going to sound paradoxical, but actually proving that 1 + 1 = 2 is way above the levels of mathematics covered on this site. If I were you, I'd take the basic laws of arithmetic as axiomatic - for now at least. If you're interested, there is a proof in Russell and Whitehead's Principia Mathematica. Be warned: it is several hundred pages of proof!
• At , shouldn't it be (k(g(x+h)-g(x))) / h, since there's no k in the denominator I don't understand how you get k((g(x+h)-g(x)) / h)?
• Does differentiable mean the derivatives of the 2 functions ARE the same or NOT the same?

For example in this question my teacher gave us a hint that we would need to find the left & right hand derivatives:
Is f differentable at x=1?
Let f(x)= 2-x if x</= 1
and (x^2)-2x+2 if x>1
(1 vote)
• how would you write rule 2 f(x)= g(x)+j(x) => f'(x) g'(x)+j'(x) in Leibniz notation?
• @ shouldn't f'(x)=0?cause when f(x)=k,f'(x)=0.Which means f'(x)=0*g(x).So when we multiply anything by zero is always zero right?
(1 vote)
• We have f(x)=k•g(x). Take the derivative of both sides to get f'(x)=d/dx (k•g(x)). So what does the right hand side simplify to?

The derivative of a product is not the product of the derivatives. That is, it's not the case that d/dx(f(x)g(x))=f'(x)g'(x). If that were the case, then every derivative would be 0, since g(x)=1•g(x). That's not useful.

Sal goes on to prove in the video why the constant gets moved outside the derivative.
• Is it just me, who's lost and confused on whats happening and don't know where to start?
(1 vote)
• It's very unlikely that it's just you.

Try watching the video again slowly. Pinpoint the exact sentence where you started feeling lost, then ask a question or go back to a previous video to try to understand that sentence.