In Session 17 at around 18:00 in lecture 6, how does the professor get $$\frac{ d }{dx }f'(kx) = k * f'(kx)$$? Was it supposed to be $$\frac{ d }{dx }f'(kx) = k * f'(x)$$? Please explain how and if this was a mistake also explain how.

Question

anonymous · Answer

No, he is correct, but he uses the chain rule in a way you normally wouldn't think to use it. For example, lets differentiate $$y = 6x$$
You know that differentiating any function in the form $$cx$$ results in $$c$$. So $$\frac{ dy }{ dx } 6x = 6$$ But lets also do this same problem using the chain rule. Let the inside function be $$u = x$$ and so  the outside function is $$y = 6u$$ Using the chain rule, $$\frac{ dy }{ dx }y = \frac{ dy }{ du }\frac{ du }{ dx }6u $$ We know $$\frac{ du }{ dx } = 1$$ So now we have $$\frac{ dy }{ dx }y = \frac{ dy }{ du }6u $$ so $$\frac{ dy }{ dx } = 6$$ Which is the same result we got above.
This is exactly what the professor is doing, using the chain rule on a constant. 
Sorry if my explanation wasn't completely clear, please ask if you still have any doubts because this is my first time answering!

anonymous · Answer

The key is to understand that the function is being applied to the constant. For example, in $$f(x)=2x^2$$ the function doesn't do anything to the constant 2, so the derivative of this function is the same as 2 times the derivative of x squared. But if this is our function, what would we get if we applied it to 3x?$$f(3x)=2(3x)^2$$Now the function is being applied to the constant, and it would be a mistake to say the derivative of this function is the same as three times the derivative of 2 times x squared. That's why we need the chain rule in this situation.