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Differentiate `ln( 1 + theta)` with respect to `sin^-1theta` .

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user8235304 | Student, Grade 11 | Valedictorian

Posted June 27, 2013 at 4:41 PM via web

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Differentiate `ln( 1 + theta)` with respect to `sin^-1theta` .

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 27, 2013 at 4:53 PM (Answer #1)

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The expression `ln(1+theta)` has to be differentiated with respect to `sin^-1theta` .

If `y = ln(1+ theta)` , `x = sin^-1 theta` and `dy/dx` has to be determined.

`dy/dx`

= `((d ln(1+ theta))/(d theta))/((d(sin^-1theta))/(d theta))`

= `(1/(theta+1) )/(1/(sqrt(1-theta^2)))`

= `sqrt(1-theta^2)/(1+theta)`

= `(sqrt(1-theta)*sqrt(1+theta))/(1+theta)`

= `sqrt(1-theta)/sqrt(1+theta)`

The required derivative `((d ln(1+ theta))/(d theta))/((d(sin^-1theta))/(d theta)) = sqrt(1- theta)/sqrt(1+theta)`

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