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,

$d/dx(tanx)^3=3(tanx)^2*d/dx(tanx)$ ,

$=3(tanx)^2(sec^2x)$ ,

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

$=3(tanx)^4+3(tanx)^2$ ,

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

$=3(tanx)^4+3(tanx)^2-3,$ as desired!