Showing dθdtanθ=11+tan2θ I suppose that dθdtanθ=darctanxdx=11+x2=11+tan2θ So is dθdtanθ=11+tan2θ correct? And dθdtanθ2=21+tan2θ2?
I suppose that
Answer & Explanation
Beginner2022-01-17Added 40 answers
It’s all a matter of dependent and independent variables, when you wrote the denominator, that , became a independent variable and became dependent variable.
But it is quite unconventional to do that, because you see you cannot obatain from without setting or or to any other thing, the key point lies in the limit definition of the derivatives. Your expression cannot be put in the form
until and unless we do some substitution for
However, your substitution is correct, and we can find the same thing by the law of derivative of inverse function,
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