why substitution with lower ranged func works in derivatives

why substitution with lower ranged func works in derivatives

Postby Guest » Tue Jun 02, 2020 10:29 pm

If I substitute x for sinx in d(x)/dx still it gives the same answer.

Why the substitution works, because the range of x is all real numbers but the range of sin is only between -1 and 1.

for more clarity i drew 2 pictures.


https://ibb.co/DfPBJkj
https://ibb.co/qksshL9
Guest
 

Re: why substitution with lower ranged func works in derivat

Postby Guest » Mon Jun 08, 2020 10:14 am

For any function "f(x)", "df/dx" is the instantaneous rate of change of f compared to the rate of change of x. Whatever "x" represents "dx/dx" is the rate of change x compared to the rate of change of x. Since those are the same, dx/dx= 1.

I am not sure what you mean by "substitute x for sin(x) in dx/dx". There is NO "sin(x)" in dx/dx to be substituted for! Perhaps you meant "substitute sin(x) for x in dx/dx".

If you meant changing dx/dx to d(sin(x))/d(sin(x)) then, yes, x changes exactly as fast as x and sin(x) changes exactly as fast as sin(x) so both derivatives are 1.

If you meant changing dx/dx to d(sin(x))/dx then d(sin(x))/dx is cos(x), not at all the same as dx/dx.
Guest
 


Return to Calculus - integrals, lim, functions



Who is online

Users browsing this forum: No registered users and 4 guests