How to show d/dx(tanx)^3=3 (tanx)^4+3(secx)^2-3?

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Answer 1 · 2018-02-26T10:23:21

Please refer to a Proof in the Explanation.

By the Chain Rule,

$\frac{d}{d x} {(tan x)}^{3} = 3 {(tan x)}^{2} \cdot \frac{d}{d x} (tan x)$ $d/dx(tanx)^3=3(tanx)^2*d/dx(tanx)$ ,

$= 3 {(tan x)}^{2} ({sec}^{2} x)$ $=3(tanx)^2(sec^2x)$ ,

$= 3 {(tan x)}^{2} {{(tan x)}^{2} + 1}$ $=3(tanx)^2{(tanx)^2+1}$ ,

$= 3 {(tan x)}^{4} + 3 {(tan x)}^{2}$ $=3(tanx)^4+3(tanx)^2$ ,

$= 3 {(tan x)}^{4} + 3 {{(sec x)}^{2} - 1}$ $=3(tanx)^4+3{(secx)^2-1}$ ,

$= 3 {(tan x)}^{4} + 3 {(tan x)}^{2} - 3,$ $=3(tanx)^4+3(tanx)^2-3,$ as desired!