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$\frac13\tan^3x-\tan x+x$

I solved it and got $\frac13\cdot3\tan x-\sec^2x+1$ by using chain rule.

I got $\tan x+\tan^3x-\tan^2x$.

The answer is $\tan^4x$. I am not able to get this.

Parcly Taxel
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Srijan
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1 Answers1

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You forgot to multiply the first term by $\sec^2x$… $$\frac d{dx}=\tan^2x(1+\tan^2x)-(1+\tan^2x)+1=\tan^4x$$

Parcly Taxel
  • 103,344
  • 1/3 * tan ^4 x.Can we use chain rule here ? – Srijan Nov 27 '20 at 07:04
  • @srijannahar that is another question, I suppose. – Parcly Taxel Nov 27 '20 at 07:07
  • Why is the first term not multiplied by tan x using chain rule. – Srijan Nov 27 '20 at 07:15
  • @srijannahar You applied the chain rule incorrectly to the first term. $\frac d{dx}\frac13\tan^3x=\frac13(3\tan^2x)\frac d{dx}\tan x$ – Parcly Taxel Nov 27 '20 at 07:16
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    $\frac{1}{3} (\tan^3 x) = \frac{1}{3} (\tan x)^3 = \frac{1}{3} (3 \tan(x)^2 \cdot \frac{d}{dx} (\tan x) )$ using the chain rule. In other words, this is $f(g(x))$ where $g(x) = \tan x$, $f(x) = x^3$, and $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$. – Toby Mak Nov 27 '20 at 07:17
  • I think you have forgot to write tan (x) in your answer. Which will be g(x). – Srijan Nov 27 '20 at 07:42