Answer to Question #204090 in Calculus for moe

Question

What is the derivative of $y = e$^arcsinx ($1/ (1+x^2)$ )

Answer 1

Given the equation

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

We can rewrite it as:

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

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