Differentiate cos^-1(2x^2-1) with respect to sin^-1√1-x^2?

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Answer 1 · 2018-01-22T05:29:54

$(df)/(dg)=sqrt((2-x^2)/(1-x^2))$

Let $f(x)=cos^(-1)(2x^2-1)$ and $g(x)=sin^(-1)(1-x^2)$

what we are seeking is $(df)/(dg)$ , which is equal to $((df)/(dx))/((dg)/(dx))$

As $f=cos^(-1)(2x^2-1)$ , we have $cosf=2x^2-1$

and differentiating $-sinf*(df)/(dx)=4x$

or $(df)/(dx)=-(4x)/sinf=-(4x)/sqrt(1-cos^2f)=-(4x)/sqrt(1-(2x^2-1))^2$

= $-(4x)/sqrt(1-4x^4+4x^2-1)=-(4x)/(2xsqrt(1-x^2))=-2/sqrt(1-x^2)$

Similarly we have $sing=1-x^2$ and $cosg*(dg)/(dx)=-2x$

or $(dg)/(dx)=-(2x)/cosg=-(2x)/sqrt(1-sin^2g)=-(2x)/sqrt(1-(1-x^2)^2)$

= $-(2x)/sqrt(1-1+2x^2-x^4)=-2/sqrt(2-x^2)$

Hence $(df)/(dg)=(-2/sqrt(1-x^2))/(-2/sqrt(2-x^2))$

= $sqrt((2-x^2)/(1-x^2))$