question based on basic diffrential calculus

[xbonez]

given,

y=x+x+x+x+........x times

calculate dy/dx

Case I : x+x+x...+5 times = 5x
x+x+x+x+.....10 times = 10x
So, x+x+x+x+....x times = xx = x^2

y=x^2
dy/dx = 2x


Case II : y=x+x+x+x+.... x times

dy/dx = d(x+x+x+x+...x times)/dx

applying sum rule,

dy/dx = d(x)/dx + d(x)/dx + d(x)/dx + ...... x times
= 1 + 1 + 1 + 1 +.....x times
= x

therefore, dy/dx = x

Please tell me which case is correct and why is the other wrong?

 

[Harvik780]

The problem with what you wrote is that x is the variable you are
differentiating with respect to x. When you write

y = x+x+x+x+...+x ,
<--x times-->

you can't say

y' = 1+1+1+1+...+1 ,
<--x times-->

because x is not constant (it's not even an integer). When you write
"x times" and find y', you are actually calculating


x *d(x)/dx = x * 1 = x ,


because it's like "taking out" one x.

Think of it this way: Why can't you say,

y = x^2 = x+x+...+x (x times)

= (1+1+...+1) (x times) \
+ (1+1+...+1) (x times) \ (x times)
+ . . . /
+ (1+1+...+1) (x times) /

so y' = 0+0+...+0 (x^2 times) = 0 ?

This time, what I actually calculated was


x^2*d(1)/dx = x^2 * 0 = 0 .


The correct way to do this is

y = x^2 = x+x+x+...+x (x times)
y' = (1+1+1+...+1) + (x) <-- 1 time!
<--x times-->
= x + x = 2x.

This is from the product rule: if f(x) = u(x)*v(x), then

f'(x) = u'(x)*v(x)+u(x)*v'(x).

So what I did was I differentiated the "x times." It's very
confusing, because what actually happened was I let u(x)=x, v(x)=x.

I hope this makes sense!

 

[xbonez]

hmm, makes sense. i had also checked it using the product rule in which case the ans comes out to be 2x