T substitution: Why is the derivative of sinx in terms of t NOT cos x in terms of t?

[hndalama]

Please see picture, for further details.

 

[ksdhart2]

The catch is that you've changed the variable. It's true that \(\displaystyle \dfrac{d(sin(x))}{dx} = \dfrac{d}{dx} \left( \dfrac{2t}{1+t^2} \right) = \dfrac{1-t^2}{1+t^2} = cos(x)\). However, \(\displaystyle \dfrac{d(sin(x))}{dt} = \dfrac{d}{dt} \left( \dfrac{2t}{1+t^2} \right) = \dfrac{2(t^2-1)}{(1+t^2)^2} \neq cos(x)\) because you're now differentiating with respect to t instead of with respect to x.

 

[sinx]

The catch is that you've changed the variable. It's true that \(\displaystyle \dfrac{d(sin(x))}{dx} = \dfrac{d}{dx} \left( \dfrac{2t}{1+t^2} \right) = \dfrac{1-t^2}{1+t^2} = cos(x)\). However, \(\displaystyle \dfrac{d(sin(x))}{dt} = \dfrac{d}{dt} \left( \dfrac{2t}{1+t^2} \right) = \dfrac{2(t^2-1)}{(1+t^2)^2} \neq cos(x)\) because you're now differentiating with respect to t instead of with respect to x.

addressing only d/dt, I have
d/dt 2t/(1+t²)= d/dt 2t(1+t²)^-1
=2(1+t²)^-1+2t(-1)(1+t²)^-2(2t)
=2/(1+t²)-4t²/(1+t²)²
=2+2t²-4t²/(1+t²)²
=2(1-t²)/(1+t²)²
you have t²-1 here,
am i wrong?

 

[Steven G]

addressing only d/dt, I have
d/dt 2t/(1+t²)= d/dt 2t(1+t²)^-1
=2(1+t²)^-1+2t(-1)(1+t²)^-2(2t)
=2/(1+t²)-4t²/(1+t²)²
=2+2t²-4t²/(1+t²)²
=2(1-t²)/(1+t²)²
you have t²-1 here,
am i wrong?

I checked your work twice and it looks fine.

 

[hndalama]

The catch is that you've changed the variable. It's true that \(\displaystyle \dfrac{d(sin(x))}{dx} = \dfrac{d}{dx} \left( \dfrac{2t}{1+t^2} \right) = \dfrac{1-t^2}{1+t^2} = cos(x)\). However, \(\displaystyle \dfrac{d(sin(x))}{dt} = \dfrac{d}{dt} \left( \dfrac{2t}{1+t^2} \right) = \dfrac{2(t^2-1)}{(1+t^2)^2} \neq cos(x)\) because you're now differentiating with respect to t instead of with respect to x.

awesome, that works. thank you

 

[sinx]

The catch is that you've changed the variable. It's true that \(\displaystyle \dfrac{d(sin(x))}{dx} = \dfrac{d}{dx} \left( \dfrac{2t}{1+t^2} \right) = \dfrac{1-t^2}{1+t^2} = cos(x)\). However, \(\displaystyle \dfrac{d(sin(x))}{dt} = \dfrac{d}{dt} \left( \dfrac{2t}{1+t^2} \right) = \dfrac{2(t^2-1)}{(1+t^2)^2} \neq cos(x)\) because you're now differentiating with respect to t instead of with respect to x.

how are you taking derivative of a f(t) with respect to x?
is there an unmentioned relation between x and t?

 

[ksdhart2]

addressing only d/dt, I have
d/dt 2t/(1+t²)= d/dt 2t(1+t²)^-1
=2(1+t²)^-1+2t(-1)(1+t²)^-2(2t)
=2/(1+t²)-4t²/(1+t²)²
=2+2t²-4t²/(1+t²)²
=2(1-t²)/(1+t²)²
you have t²-1 here,
am i wrong?

Whoops! Yes, I missed a minus sign out front of the whole thing. It should really be \(\displaystyle -\dfrac{2(t^2-1)}{(1+t^2)^2}\). Flipping the order of the terms in the numerator would achieve the same effect. Sorry about my mistake.

how are you taking derivative of a f(t) with respect to x?
is there an unmentioned relation between x and t?

In OP's picture, they note the relationship \(\displaystyle sin(x) = \dfrac{2t}{1+t^2}\), which implies that there was a t-substitution of \(\displaystyle t=tan \left( \dfrac{x}{2} \right)\) (alternatively, \(\displaystyle t=cot \left( \dfrac{x}{2} \right)\) would also work)

 

[sinx]

Whoops! Yes, I missed a minus sign out front of the whole thing. It should really be \(\displaystyle -\dfrac{2(t^2-1)}{(1+t^2)^2}\). Flipping the order of the terms in the numerator would achieve the same effect. Sorry about my mistake.



In OP's picture, they note the relationship \(\displaystyle sin(x) = \dfrac{2t}{1+t^2}\), which implies that there was a t-substitution of \(\displaystyle t=tan \left( \dfrac{x}{2} \right)\) (alternatively, \(\displaystyle t=cot \left( \dfrac{x}{2} \right)\) would also work)

now the sign makes sense.
I see the sin(x)=2t/(1+t²) and am aware this implies t=tan(x/2) now.
I will try to work it from here.
thanks.

 

[Steven G]

how are you taking derivative of a f(t) with respect to x?
is there an unmentioned relation between x and t?

Actually there is a mention relation between x and t and that is sin(x)= 2t/(1+t²). Is this what you mean??

 

[sinx]

Actually there is a mention relation between x and t and that is sin(x)= 2t/(1+t²). Is this what you mean??

I was asking how do you take the derivative of 2t/(1+t²) with respect to x?

 

[Steven G]

I was asking how do you take the derivative of 2t/(1+t²) with respect to x?

You probably will not like my answer but it is what it is.

Since sin(x) = 2t/(1+t²) we have (d/dx) [2t/(1+t²)] = (d/dx)sinx =cosx.

 

[sinx]

You probably will not like my answer but it is what it is.

Since sin(x) = 2t/(1+t²) we have (d/dx) [2t/(1+t²)] = (d/dx)sinx =cosx.

you are right, i did not like that answer.
I am aware that d/dx sinx=cosx, but it does not give cos(t), or cosx=f(t).
I expected a route such as dsinx/dx=dsinx/dt*dt/dx, ....chain rule, or something like that.

from an earlier post, sinx=2t/[1+t²] implies (and it can be shown) that t=tan(^x/₂).
then sinx=2tan(^x/₂)/[1+tan²(^x/₂)], or some combination, such as 2cos(^x/₂)sin(^x/₂), that you can take the derivative of with respect to x.
then d/dx= cos²(x/2)-sin²(x/2)=cos2(x/2)=cosx.
that is the long way around, but i think is the purpose of the exercise.

 

[Steven G]

you are right, i did not like that answer.
I am aware that d/dx sinx=cosx, but it does not give cos(t), or cosx=f(t).
I expected a route such as dsinx/dx=dsinx/dt*dt/dx, ....chain rule, or something like that.

from an earlier post, sinx=2t/[1+t²] implies (and it can be shown) that t=tan(^x/₂).
then sinx=2tan(^x/₂)/[1+tan²(^x/₂)], or some combination, such as 2cos(^x/₂)sin(^x/₂), that you can take the derivative of with respect to x.
then d/dx= cos²(x/2)-sin²(x/2)=cos2(x/2)=cosx.
that is the long way around, but i think is the purpose of the exercise.

y = 2t/[1+t²]=>dy/dt = 2(1-t²)/(1+t²)²

now t= tan (x/2). Draw a right triangle that corresponds to t= tan (x/2). So opposite x/2 is t and adjacent to x/2 is 1. So the hypotenuse is sqrt(1+t²). Now we have the complete triangle.

Since t= tan (x/2) we get dt/dx = sec² (x/2)*(1/2)

Now dy/dx = (dy/dt)(dt/dx) = 2(1-t²)/(1+t²)²*sec² (x/2)*(1/2)=(1-t²)/(1+t²)²*sec² (x/2). Now use the triangle to compute sec² (x/2) in terms of t. Multiply this result by 2(1-t²)/(1+t²)² and you get the desired result.

 

[sinx]

y = 2t/[1+t²]=>dy/dt = 2(1-t²)/(1+t²)²

now t= tan (x/2). Draw a right triangle that corresponds to t= tan (x/2). So opposite x/2 is t and adjacent to x/2 is 1. So the hypotenuse is sqrt(1+t²). Now we have the complete triangle.

Since t= tan (x/2) we get dt/dx = sec² (x/2)*(1/2)

Now dy/dx = (dy/dt)(dt/dx) = 2(1-t²)/(1+t²)²*sec² (x/2)*(1/2)=(1-t²)/(1+t²)²*sec² (x/2). Now use the triangle to compute sec² (x/2) in terms of t. Multiply this result by 2(1-t²)/(1+t²)² and you get the desired result.

very good.
more than one way to skin a cat.
thanks