Can anyone help me differentiate this function? y=sec x/ 1+ sec x

2 Answers
Feb 26, 2018

#sin x/(1+cos x)^2#

Explanation:

#y= secx/{1+sec x} = 1/{1+cos x}#

So

#dy/dx = -1/(1+cosx)^2 times (-sin x) = sin x/(1+cos x)^2#

Feb 26, 2018

Given:

#y = sec(x)/(1+sec(x)#

We can convert the secant functions into a single cosine function, if we multiply the fraction by 1 in the form of #cos(x)/cos(x)#:

#y = cos(x)/cos(x)sec(x)/(1+sec(x)#

Performing the multiplication:

#y = (cos(x)sec(x))/(cos(x)+cos(x)sec(x)#

We eliminate both secant functions, using the identity #cos(x)sec(x) = 1#:

#y = 1/(cos(x)+1)#

To differentiate, use the quotient rule,

#(d((u(x))/(v(x))))/dx = ((d(u(x)))/dxv(x)-u(x)(d(v(x)))/dx)/(v(x))^2#

where #u(x) = 1# and #v(x) = cos(x)+1#:

#(d((1)/(cos(x)+1)))/dx = ((d(1))/dx(cos(x)+1)-1(d(cos(x)+1))/dx)/(cos(x)+1)^2#

The first term in the numerator is 0, because the derivative of a constant is 0:

#(d((1)/(cos(x)+1)))/dx = (-1(d(cos(x)+1))/dx)/(cos(x)+1)^2#

The derivative in the remaining term is #-sin(x)#

#(d((1)/(cos(x)+1)))/dx = sin(x)/(cos(x)+1)^2#