Posted by Charles on Sunday, September 28, 2008 at 10:02pm.
recall that d(secu)/dx
= tanu secu du/dx
so if f(x) = 3sec^2(pix-1)
= 3(sec(pix-1))^2 then using the chain rule
f'(x) = 6(sec(pix-1)*tan(pix-1)sec(pix-1)*pi
= 6pi(sin(pix-1)/cos(pix-1)*sec^2(pix-1)
= 6pi(sin(pix-1))/cos(pix-1)*1/cos^2(pix-1)
= 6pi(sinpix-1)/cos^3(pix-1)
Related Questions
calculus - Differentiate both sides of the double angle identify sin 2x= 2 sin x...
calc - find the area between the x-axis and the graph of hte given function over...
Calculus - How would I find the antiderivative of this equation? S 0 to pi/3 (1-...
Calculus (Derivatives of Inverse Functions) - Suppose f(x) = sin(pi cos(x)). On ...
calc - d/dx integral from o to x of function cos(2*pi*x) du is first i do the ...
calculus - The limit represents the derivative of some function f at some number...
calculus - Could someone check my reasoning? thanx Find the derivative of the ...
please help me calc. have test tom - d/dx integral from o to x of function cos(2...
calculus =) - find the derivative of g(x)=sin(cos(sin(pi*x)))
Math(Please help) - 1)tan Q = -3/4 Find cosQ -3^2 + 4^2 = x^2 9+16 = sqrt 25 = 5...
For Further Reading
