Question

Find the differentiation of the following function

tan3xtan2xtanx with respect to x.

Don’t use product rule and logarithms

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: \: tan3x \: tan2x \: tanx$

Let we first represent this product of terms in form of addition or subtraction with the help of Trigonometric functions.

We know,

$\rm :\longmapsto\:3x = 2x + x$

So,

$\rm :\longmapsto\:tan3x = tan(2x + x)$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany}}}}$

So, using this identity, we get

$\rm :\longmapsto\:tan3x = \dfrac{tan2x + tanx}{1 - tan2x \: tanx}$

$\rm :\longmapsto\:tan3x - tan3xtan2xtanx = tan2x + tanx$

$\rm\implies \:tan3xtan2xtanx = tan3x - tan2x - tanx$

So,

Let assume that

$\rm :\longmapsto\:y = tan3xtan2xtanx$

can be rewritten as

$\rm :\longmapsto\:y = tan3x - tan2x - tanx$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}( tan3x - tan2x - tanx)$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}tanx = {sec}^{2}x \: }}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {sec}^{2}3x\dfrac{d}{dx}3x - {sec}^{2}2x\dfrac{d}{dx}2x - {sec}^{2}x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 3{sec}^{2}3x - 2{sec}^{2}2x - {sec}^{2}x$

Hence,

$\rm\implies \:\boxed{\tt{ \:\dfrac{dy}{dx} = 3{sec}^{2}3x - 2{sec}^{2}2x - {sec}^{2}x}}$

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MORE TO KNOW

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

Answer:

$\boxed{\huge{\mathbb\pink{REFERR \:TO\: THE\:\: ATTACHMENT }}}$

Step-by-step explanation:

hope it help you.

thanks