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

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Answer 1 · 2017-10-03T14:38:01

#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 !!!!!

Answer 2 · 2017-10-03T14:42:26

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#