Answer to Question #114646 in Calculus for ANJU JAYACHANDRAN

Question #114646
Find the derivative of In(1+x^2) w.r.t tan inverse x
1
Expert's answer
2020-05-20T18:41:19-0400

Let us assume that:

y=ln(1+x2)y = \ln (1 + x^2)

And:

v=tan1(x)v = \tan^{-1} (x)

Derivative of yy with respect to xx is:

dydx=ddx(ln(1+x2))=11+x2ddx(1+x2)\frac{dy}{dx} = \frac{d}{dx} \left( \ln (1 + x^2) \right) \\ \hspace{0.5 cm} = \frac{1}{1+x^2} \frac{d}{dx} (1 + x^2) \\

=11+x2[ddx(1)+ddx(x2)]\hspace{0.5 cm} = \frac{1}{1 + x^2} \left[ \frac{d}{dx} (1) + \frac{d}{dx} (x^2) \right]

=11+x2(0+2x)\hspace{0.5 cm} = \frac{1}{1+x^2} \left( 0 + 2x \right) \hspace{1 cm} [ ddx(xn)=nxn1andddx(c)=0\because \frac{d}{dx} (x^n) = nx^{n-1} \, and \, \frac{d}{dx} (c) = 0 \, ( Where c is a constant ) ]

=2x1+x2\hspace{0.5 cm} = \frac{2x}{1 + x^2}

Derivative of vv with respect to xx is:

dvdx=ddx(tan1(x))=11+x2\frac{dv}{dx} = \frac{d}{dx} \left( \tan^{-1} (x) \right) \\ \hspace{0.5 cm} = \frac{1}{1+x^2} [ ddx(tan1(x))=11+x2\because \frac{d}{dx} \left( \tan^{-1} (x) \right) = \frac{1}{1+x^2} ]

Now:

dydv=dydxdvdx=2x1+x211+x2=2x\frac{dy}{dv} = \frac{\frac{dy}{dx}}{\frac{dv}{dx}} = \frac{\frac{2x}{1+x^2}}{\frac{1}{1+x^2}} = 2x

Hence:

The derivative of ln(1+x2)\ln (1 + x^2) with respect to tan1(x)\tan^{-1} (x) or derivative of yy with respect to vv is:

dydv=2x\frac{dy}{dv} = 2x


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