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 $\displaystyle y=\frac{e^{tan^{-1}x}}{\sqrt{1+3x}}$

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- Sep 9th 2011, 08:55 AMPunchequation of tangent
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 $\displaystyle y=\frac{e^{tan^{-1}x}}{\sqrt{1+3x}}$

- Sep 9th 2011, 09:22 AMalexmahoneRe: equation of tangent
Actually, $\displaystyle e^x=1+x+\frac{1}{2}x^2+...$

Quote:

deduce the equation of the tangent to the curve $\displaystyle y=\frac{e^{tan^{-1}x}}{\sqrt{1+3x}}$

- Sep 11th 2011, 04:19 AMPunchRe: equation of tangent
- Sep 11th 2011, 04:26 AMalexmahoneRe: equation of tangent
- Sep 13th 2011, 02:43 AMPunchRe: equation of tangent
- Sep 13th 2011, 03:36 AMalexmahoneRe: equation of tangent