How can I find the derivative of y = tan(x)?
Let's first recall the definition of tan(x) = sin(x)/cos(x). Hence, y' = d/dx( sin(x)/cos(x) ). Recall quotient rule for differentiation: ( f(x)/g(x) )' = ( f '(x)^2*g(x) - f(x)*g'(x)^2 ) / g(x)^2 and that d/dx( sin(x) ) = cos(x), and d/dx( cos(x) ) = - sin(x). Then, y' = ( cos(x)cos(x) - sin(x)(-sin(x)) )/cos^2(x) => y' = ( cos^2(x) + sin^2(x) ) /cos^2(x). Recalling that cos^2(x) + sin^2(x) = 1 by the Pythagorean trigonometric identity, then y' = 1 / cos^2(x). Since sec(x) = 1/cos(x), we can state our result as: y' = d/dx( tan(x) ) = sec^2(x).
RB
Related Maths A Level answers
All answers ▸Express 4sin(x)+6cos(x) in terms of Rsin(x+a) where R and a are constants to be determined (a should be given in rad).
How does integration by parts work ad when to use it?
The curve C has equation y = 3x^4 – 8x^3 – 3 (a) Find (i) dy/dx (ii) d^2y/dx^2 (3 marks) (b) Verify that C has a stationary point when x = 2 (2marks) (c) Determine the nature of this stationary point, giving a reason for your answer. (2)
y = 2x^3 + 15x^2 + 24x + 10 Find the stationary points on this curve and determine their nature