Raphael K. answered • 09/17/21
I genuinely love teaching Calculus and have for 10+ years.
Find the derivative of f(θ) = cos√(sin(tan(πθ))). Do not leave negative exponents.
FUN! product rule with the 1/2 power then chain rule with the d/dx of sin and tan functions....
d/dx[cos1/2(sin(tan(nθ)))]
= 1/2*[cos-1/2(sin(tan(nθ))]*cos(tan(nθ))*(n*sec2(nθ))dθ
or
[cos(tan(nθ))*(n*sec2(nθ))] / 2[cos1/2(sin(tan(nθ))]
Raphael K.
It is correct, where the 1/2 power is.09/17/21
Raphael K.
the answer is 1/2 cos to the -1/2 of sin of tan n theta, times negative sin of sin of tan n theta times cos of tan n theta times n secant squared of n theta.09/17/21
Raphael K.
[-sin(sin(tan(n0)) * cos(tan(n0)) *n * sec^2(n0) ] / [cos^(1/2)(sin(tan(n0))]09/17/21
Raphael K.
I might have missed the first cos derivative term after the power rule.09/17/21
Raphael K.
[-sin(sin(tan(n0)) * cos(tan(n0)) *n * sec^2(n0) ] / 2[cos^(1/2)(sin(tan(n0))]09/17/21
Raphael K.
i forgot the 2 in the denominator. silly problem09/17/21
Noah S.
Wouldn't that be incorrect because of where the 1/2 power is? For example, cos²(x) ≠ cos(x²)09/17/21