Showing dθdtan⁡θ=11+tan2θ I suppose that dθdtan⁡θ=darctan⁡xdx=11+x2=11+tan2θ So is dθdtan⁡θ=11+tan2θ correct? And dθdtan⁡θ2=21+tan2θ2?

William Boggs

William Boggs

Answered question

2022-01-16

Showing dθdtanθ=11+tan2θ
I suppose that
dθdtanθ=darctanxdx=11+x2=11+tan2θ
So is
dθdtanθ=11+tan2θ
correct? And
dθdtanθ2=21+tan2θ2?

Answer & Explanation

Paul Mitchell

Paul Mitchell

Beginner2022-01-17Added 40 answers

It’s all a matter of dependent and independent variables, when you wrote dθdtanθ the denominator, that tanθ, became a independent variable and θ became dependent variable.
But it is quite unconventional to do that, because you see you cannot obatain θ from tanθ without setting tanθ=x or tanθ=a or to any other thing, the key point lies in the limit definition of the derivatives. Your expression dθdtanθ cannot be put in the form
limh0f(x+h)f(x)h
until and unless we do some substitution for tanθ
However, your substitution is correct, and we can find the same thing by the law of derivative of inverse function,
ddxf1(x)=(df(x)dx)1
levurdondishav4

levurdondishav4

Beginner2022-01-18Added 38 answers

Heres

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