How do you differentiate #y=sec^-1(2x)+csc^-1(2x)#?

Question

Christopher Agresta · Accepted Answer

#0#

#d/dxsec^(-1)x = 1/(|quadx|sqrt(x^2 - 1))#

#d/dxcosec^(-1)x = -1/(|quadx|sqrt(x^2 - 1))#

#therefore#, #y = sec^-1(2x) + cosec^-1(2x)#

#dy/dx = 1/(|quad2x|sqrt(4x^2 - 1)) * 2 - 1/(|quad2x|sqrt(4x^2 - 1)) * 2#

Hence, #dy/dx = 0#

ENJOY MATHS !!!!!

Charles Anctil · Answer

Differentiate each term, using table lookup and the chain rule.
Combine the terms.

Given: #y=sec^-1(2x)+csc^-1(2x)# Differentiate each term: #dy/dx=(d(sec^-1(2x)))/dx+(d(csc^-1(2x)))/dx" [1]"# From any reference table, we know that: #(d(sec^-1(u)))/(du) = 1/(|u|sqrt(u^2-1)# In our case, #u = 2x#, then #(du)/dx = 2# Using the chain rule: #(d(sec^-1(2x)))/(dx) = 2/(|2x|sqrt(4x^2-1))" [2]"# Similarly, we know that: #(d(csc^-1(u)))/(du) = -1/(|u|sqrt(u^2-1)# In our case, #u = 2x#, then #(du)/dx = 2# Using the chain rule: #(d(csc^-1(2x)))/(dx) = -2/(|2x|sqrt(4x^2-1))" [3]"# Substitute equations [2] and [3] into equation [1]: #dy/dx=2/(|2x|sqrt(4x^2-1))-2/(|2x|sqrt(4x^2-1))# The terms sum to 0: #dy/dx= 0#

Charles Arocha · Answer

undefined To differentiate $ y = \sec^{-1}(2x) + \csc^{-1}(2x) $:

1. Use the chain rule for differentiation.
2. Differentiate each term separately.
3. For $ \sec^{-1}(2x) $, differentiate $ \sec^{-1}(u) $ with respect to $ u $ and then multiply by the derivative of $ u = 2x $.
4. For $ \csc^{-1}(2x) $, differentiate $ \csc^{-1}(v) $ with respect to $ v $ and then multiply by the derivative of $ v = 2x $.

The derivatives of the inverse trigonometric functions are as follows:

- $ \frac{d}{du}(\sec^{-1}u) = \frac{1}{|u|\sqrt{u^2-1}} $
- $ \frac{d}{dv}(\csc^{-1}v) = -\frac{1}{|v|\sqrt{v^2-1}} $

Apply the chain rule and simplify the derivatives accordingly.