Answer to Question #204090 in Calculus for moe

Question #204090

What is the derivative of y=ey = earcsinx (1/(1+x2)1/ (1+x^2) )


1
Expert's answer
2021-06-09T14:30:18-0400

Given the equation


y=earcsinx(11+x2)y= e^{arcsinx} \Big( \frac{1}{1+x^2}\Big)

We can rewrite it as:


y=earcsinx1+x2y= \frac{e^{arcsinx}}{1+x^2}

ddx[earcsin(x)x2+1]=ddx[earcsin(x)](x2+1)earcsin(x)ddx[x2+1](x2+1)2=earcsin(x)ddx[arcsin(x)](x2+1)earcsin(x)(ddx[x2]+ddx[1])(x2+1)2=(x2+1)earcsin(x)1x22xearcsin(x)(x2+1)2 Therefore the derivative is dydx=earcsin(x)1x2(x2+1)2xearcsin(x)(x2+1)2\frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{\mathrm{e}^{\arcsin (x)}}{x^{2}+1}\right]\\ =\frac{\frac{\mathrm{d}}{\mathrm{d} x}\left[\mathrm{e}^{\arcsin (x)}\right] \cdot\left(x^{2}+1\right)-\mathrm{e}^{\arcsin (x)} \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left[x^{2}+1\right]}{\left(x^{2}+1\right)^{2}}\\ =\frac{\mathrm{e}^{\arcsin (x)} \cdot \frac{\mathrm{d}}{\mathrm{d} x}[\arcsin (x)] \cdot\left(x^{2}+1\right)-\mathrm{e}^{\arcsin (x)}\left(\frac{\mathrm{d}}{\mathrm{d} x}\left[x^{2}\right]+\frac{\mathrm{d}}{\mathrm{d} x}[1]\right)}{\left(x^{2}+1\right)^{2}}\\ =\frac{\frac{\left(x^{2}+1\right) \mathrm{e}^{\mathrm{arcsin}(x)}{\sqrt{1-x^{2}}}}-2 x \mathrm{e}^{\arcsin (x)}}{\left(x^{2}+1\right)^{2}}\\ \text{ Therefore the derivative is }\\ \dfrac{dy}{dx}=\frac{\mathrm{e}^{\arcsin (x)}}{\sqrt{1-x^{2}}\left(x^{2}+1\right)}-\frac{2 x \mathrm{e}^{\arcsin (x)}}{\left(x^{2}+1\right)^{2}}


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