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find the equation of a line tangent to g(x)=4x*sin(2x) at x=pi/8
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slope of g(x) at x=pi/8 is given by the derivative at the point x=pi/8. At x=pi/8, derivative of g(x) is 4sin(2x)+8xcos(2x) = (4+pi)/(2^1/2) Slope of the tangent is this. Now, it passes through the point (pi/8,pi/(2*2^1/2)) (y-(pi/2*2^1/2))/(x-pi/8)=(4+pi)/(2^1/2)
oh so u use product rule...i was using chain rule and gettin all confused lol. thanks =)
Yes product rule. :)
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