how do you differentiate tan(x)

The differential of tan(x) is sec2(x), however knowing how dy/dx tan(x)=sec2(x) is very useful in differentiating other trigonometric functions. So, if we first employ the rule that tan(x)=sin(x)/cos(x), which we can then differentiate using the quotient rule. The quotient rule is (VU'-UV')/V2 where V' and U' are the differentials of the values of V and U.If we let U=sin(x) and V=cos(x), then U'=cos(x) and V'= -sin(x). The quotient rule is (VU'-UV')/V2, so if we sub in our values we get (cos(x)cos(x))-(sin(x)-sin(x))/ Cos2(x), simplified down we get (cos2(x) - (-sin2(x))/cos2(x). As we know, when you take away a negative, it becomes positive, so cos2(x) - (-sin2(x)) becomes cos2(x) + sin2(x), and fourtunately there is a trigonometric rule which states that cos2(x) + sin2(x) = 1. Using this we go back to our original equation and we have 1/cos2(x), which is equivalent to sec2(x), thus showing how we derive the differential of tan(x)

OS

Related Maths A Level answers

All answers ▸

What is the y-coordinate minimum point of y = 3x^2 + x - 4


How do I break down (x-2)/((x+1)(x-1)^2) into partial fractions?


Rationalise the denominator of 25/sqrt(5)


How would you solve the inequality x^2-2x-8 >= 0?