How do you differentiate $f (x) = ln (tan x - x^{2})$ $f(x)= ln(tanx-x^2)$ ?

Question

1604 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-11T20:34:12

$\frac{{sec}^{2} x - 2 x}{tan x - x^{2}}$ $(sec^2x-2x)/(tanx-x^2)$

Set $y = ln u, u = tan x - x^{2}$ $y=lnu, u=tanx-x^2$

$\frac{d y}{d u} = \frac{1}{u}, \frac{d u}{d x} = {sec}^{2} x - 2 x$ $dy/(du)=1/u, (du)/(dx)=sec^2x-2x$

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)*(du)/(dx)$

$= \frac{1}{tan x - x^{2}} \cdot {sec}^{2} x - 2 x$ $=1/(tanx-x^2)*sec^2x-2x$

$= \frac{{sec}^{2} x - 2 x}{tan x - x^{2}}$ $=(sec^2x-2x)/(tanx-x^2)$

How do you differentiate f(x)=ln(tanx−x2)f(x)= ln(tanx-x^2) ?