Given that $\displaystyle e^{tan^{-1}x}=1+x+\frac{1}{2}x^2+...$ and $\displaystyle (1+3x)^{-\frac{1}{2}}=1-\frac{3}{2}x+\frac{27}{8}x^2+..., $deduce the equation of the tangent to the curve $\displaystyle y=\frac{e^{tan^{-1}x}}{\sqrt{1+3x}}$
Given that $\displaystyle e^{tan^{-1}x}=1+x+\frac{1}{2}x^2+...$ and $\displaystyle (1+3x)^{-\frac{1}{2}}=1-\frac{3}{2}x+\frac{27}{8}x^2+..., $deduce the equation of the tangent to the curve $\displaystyle y=\frac{e^{tan^{-1}x}}{\sqrt{1+3x}}$
Actually, $\displaystyle e^x=1+x+\frac{1}{2}x^2+...$
Write it as $\displaystyle y=(1+3x)^{-\frac{1}{2}}e^{tan^{-1}x}$ and use the product rule. (Make sure you use the correct formula for $\displaystyle e^{tan^{-1}x}$.)deduce the equation of the tangent to the curve $\displaystyle y=\frac{e^{tan^{-1}x}}{\sqrt{1+3x}}$