Answer
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Hint: Students should use CHAIN RULE while solving this problem. Derivative of \[\tan x\] is \[{{\sec }^{2}}x\] i.e \[\dfrac{d}{dx}\tan a={{\sec }^{2}}a\dfrac{da}{dx}\] .Students should know all the derivative differentiation formulas for solving this kind of questions.
Complete step-by-step solution:
By reading the question carefully we came to know that we have to find the derivative of \[{{\tan }^{2}}(3x)\].
So let us consider given equation \[{{\tan }^{2}}(3x)\] as \[y\]
So \[y={{\tan }^{2}}(3x)\]…………eq(1)
Now it's clear that we have to find \[\dfrac{dy}{dx}\]. Where \[y={{\tan }^{2}}(3x)\].
So let us proceed the calculation by multiplying POWERS by \[\dfrac{1}{2}\] on both sides i.e LHS & RHS.
So we get the equation as
\[{{y}^{1\times \dfrac{1}{2}}}={{\tan }^{2\times \dfrac{1}{2}}}(3x)\].
On simplification we get,
\[{{y}^{\dfrac{1}{2}}}=\tan (3x)\]
Now let us differentiate the whole equation with respect to \[y\]. i.e we have to differentiate both LHS & RHS with respective to \[y\],
So we get equation as
\[\dfrac{d}{dx}{{y}^{\dfrac{1}{2}}}=\dfrac{d}{dx}\tan (3x)\]
We know from the basic derivative formulas that \[\dfrac{d}{dx}{{a}^{n}}=n\times {{a}^{n-1}}\dfrac{da}{dx}\] and \[\dfrac{d}{dx}\tan a={{\sec }^{2}}a\dfrac{da}{dx}\].
So now we can write \[\dfrac{d}{dx}{{y}^{\dfrac{1}{2}}}=\dfrac{1}{2}{{y}^{-\dfrac{1}{2}}}\dfrac{dy}{dx}\] and \[\dfrac{d}{dx}\tan (3x)={{\sec }^{2}}(3x)\dfrac{d}{dx}3x\]
So we get new equation as
\[\dfrac{1}{2}{{y}^{\dfrac{1}{2}-1}}\dfrac{dy}{dx}={{\sec }^{2}}(3x)\dfrac{d}{dx}3x\]
from the basic derivative formulas that \[\dfrac{d}{dx}ka=k\times \dfrac{da}{dx}\] where \[k=\]constant
so now we can write
\[\dfrac{d}{dx}(3x)=3\times \dfrac{dx}{dx}\]
\[\dfrac{d}{dx}(3x)=3\times 1\] where \[\dfrac{dx}{dx}=1\]
So now we get new equation as
\[\dfrac{1}{2}{{y}^{-\dfrac{1}{2}}}\dfrac{dy}{dx}={{\sec }^{2}}(3x)\times 3\times \dfrac{dx}{dx}\]
On simplification we get equation as
\[\dfrac{1}{2}\dfrac{1}{\sqrt{y}}\dfrac{dy}{dx}={{\sec }^{2}}(3x)\times 3\times 1\]
Now multiply with \[2\sqrt{y}\] on both sides
Now we will equation as
\[2\sqrt{y}\times \dfrac{1}{2}\dfrac{1}{\sqrt{y}}\dfrac{dy}{dx}={{\sec }^{2}}(3x)\times 3\times 2\sqrt{y}\]…………..(2)
On simplification we get
From eq(1) we know that \[y={{\tan }^{2}}(3x)\]
Now let's replace \[y={{\tan }^{2}}(3x)\] in eq(2)
So,
\[\dfrac{dy}{dx}={{\sec }^{2}}(3x)\times 6\times \sqrt{{{\tan }^{2}}\left( 3x \right)}\]
From the basic formula we know that \[\sqrt{{{x}^{2}}}=x\] we can write \[\sqrt{{{\tan }^{2}}\left( 3x \right)}=\tan (3x)\]
So now we get new equation as
\[\dfrac{dy}{dx}={{\sec }^{2}}(3x)\times 6\times \tan (3x)\]
On arranging terms we get the final answer as
\[\dfrac{dy}{dx}=6\times {{\sec }^{2}}(3x)\times \tan (3x)\].
Note: Students should know all basic derivative formulas. We should try to avoid calculation mistakes because small mistakes in calculations can lead to major errors. Also, the possible mistake we can make while solving the question is misreading the question \[{{\tan }^{2}}(3x)\] as \[\tan {{(3x)}^{2}}\] and then solving it wrong.
Complete step-by-step solution:
By reading the question carefully we came to know that we have to find the derivative of \[{{\tan }^{2}}(3x)\].
So let us consider given equation \[{{\tan }^{2}}(3x)\] as \[y\]
So \[y={{\tan }^{2}}(3x)\]…………eq(1)
Now it's clear that we have to find \[\dfrac{dy}{dx}\]. Where \[y={{\tan }^{2}}(3x)\].
So let us proceed the calculation by multiplying POWERS by \[\dfrac{1}{2}\] on both sides i.e LHS & RHS.
So we get the equation as
\[{{y}^{1\times \dfrac{1}{2}}}={{\tan }^{2\times \dfrac{1}{2}}}(3x)\].
On simplification we get,
\[{{y}^{\dfrac{1}{2}}}=\tan (3x)\]
Now let us differentiate the whole equation with respect to \[y\]. i.e we have to differentiate both LHS & RHS with respective to \[y\],
So we get equation as
\[\dfrac{d}{dx}{{y}^{\dfrac{1}{2}}}=\dfrac{d}{dx}\tan (3x)\]
We know from the basic derivative formulas that \[\dfrac{d}{dx}{{a}^{n}}=n\times {{a}^{n-1}}\dfrac{da}{dx}\] and \[\dfrac{d}{dx}\tan a={{\sec }^{2}}a\dfrac{da}{dx}\].
So now we can write \[\dfrac{d}{dx}{{y}^{\dfrac{1}{2}}}=\dfrac{1}{2}{{y}^{-\dfrac{1}{2}}}\dfrac{dy}{dx}\] and \[\dfrac{d}{dx}\tan (3x)={{\sec }^{2}}(3x)\dfrac{d}{dx}3x\]
So we get new equation as
\[\dfrac{1}{2}{{y}^{\dfrac{1}{2}-1}}\dfrac{dy}{dx}={{\sec }^{2}}(3x)\dfrac{d}{dx}3x\]
from the basic derivative formulas that \[\dfrac{d}{dx}ka=k\times \dfrac{da}{dx}\] where \[k=\]constant
so now we can write
\[\dfrac{d}{dx}(3x)=3\times \dfrac{dx}{dx}\]
\[\dfrac{d}{dx}(3x)=3\times 1\] where \[\dfrac{dx}{dx}=1\]
So now we get new equation as
\[\dfrac{1}{2}{{y}^{-\dfrac{1}{2}}}\dfrac{dy}{dx}={{\sec }^{2}}(3x)\times 3\times \dfrac{dx}{dx}\]
On simplification we get equation as
\[\dfrac{1}{2}\dfrac{1}{\sqrt{y}}\dfrac{dy}{dx}={{\sec }^{2}}(3x)\times 3\times 1\]
Now multiply with \[2\sqrt{y}\] on both sides
Now we will equation as
\[2\sqrt{y}\times \dfrac{1}{2}\dfrac{1}{\sqrt{y}}\dfrac{dy}{dx}={{\sec }^{2}}(3x)\times 3\times 2\sqrt{y}\]…………..(2)
On simplification we get
From eq(1) we know that \[y={{\tan }^{2}}(3x)\]
Now let's replace \[y={{\tan }^{2}}(3x)\] in eq(2)
So,
\[\dfrac{dy}{dx}={{\sec }^{2}}(3x)\times 6\times \sqrt{{{\tan }^{2}}\left( 3x \right)}\]
From the basic formula we know that \[\sqrt{{{x}^{2}}}=x\] we can write \[\sqrt{{{\tan }^{2}}\left( 3x \right)}=\tan (3x)\]
So now we get new equation as
\[\dfrac{dy}{dx}={{\sec }^{2}}(3x)\times 6\times \tan (3x)\]
On arranging terms we get the final answer as
\[\dfrac{dy}{dx}=6\times {{\sec }^{2}}(3x)\times \tan (3x)\].
Note: Students should know all basic derivative formulas. We should try to avoid calculation mistakes because small mistakes in calculations can lead to major errors. Also, the possible mistake we can make while solving the question is misreading the question \[{{\tan }^{2}}(3x)\] as \[\tan {{(3x)}^{2}}\] and then solving it wrong.