Home > Calculus > Derivatives of y=sec(x), y=cot(x), y= csc(x)

What is the derivative of #y=ln(sec(x)+tan(x))#?

Answer 1

Ezekiel Alexander

Answer: #y'=sec(x)#

Full explanation:

Suppose, #y=ln(f(x))#

Using chain rule, #y'=1/f(x)*f'(x)#

Similarly, if we follow for the problem, then

#y'=1/(sec(x)+tan(x))*(sec(x)+tan(x))'#

#y'=1/(sec(x)+tan(x))*(sec(x)tan(x)+sec^2(x)) #

#y'=1/(sec(x)+tan(x))*sec(x)(sec(x)+tan(x))#

#y'=sec(x)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Cameron Alexander

Will give you a personal video explanation of how it's done...

Learn how to differentiate y=ln(secx+tanx) in this video

Alternatively, you can use these workings...

#ln(secx+tanx)=y#

#e^y=secx+tanx#

#e^y*(dy)/(dx)=secxtanx+sec^2x#

#e^y*(dy)/(dx)=secx(secx+tanx)#

#(dy)/(dx)=(secx(secx+tanx))/e^y#

#(dy)/(dx)=(secx(secx+tanx))/((secx+tanx))#

#(dy)/(dx)=secx#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 3

Christopher Allers

To find the derivative of ( y = \ln(\sec(x) + \tan(x)) ), you would use the chain rule. The derivative is:

[ \frac{d}{dx}\ln(\sec(x) + \tan(x)) = \frac{1}{\sec(x) + \tan(x)} \cdot (\sec(x)\tan(x) + \sec^2(x)) ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.