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

2 Answers
Rutik Tupe
Oct 3, 2017

0

Explanation:

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 !!!!!

Douglas K.
Oct 3, 2017

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

Explanation:

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

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