Question

How do you find the derivative of \[y = \tan (2x)\] ?

Answer 1

Hint: We know that tangent is the ratio of sine and cosine function. That is \[\tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}}\] . To solve this we use the quotient rule that, if the two functions \[f(x)\] and \[g(x)\] are differentiable then the quotient is differentiable and \[{\left( {\dfrac{{f(x)}}{{g(x)}}} \right)^1} = \dfrac{{f'(x)g(x) - f(x)g'(x)}}{{{{\left( {g(x)} \right)}^2}}}\] .

Complete step-by-step answer:
Given, \[y = \tan (2x)\] .
We need to find the differentiation of \[\tan (2x)\] .
We know from the definition of tangent we have \[\tan (2x) = \dfrac{{\sin (2x)}}{{\cos (2x)}}\] .
So we have,
 \[ \Rightarrow y = \dfrac{{\sin (2x)}}{{\cos (2x)}}\] .
Now differentiating with respect to ‘x’, we have
 \[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{\sin (2x)}}{{\cos (2x)}}} \right)\]
We know that \[\sin (2x)\] and \[\cos (2x)\] are differentiable. Now we have quotient rule \[{\left( {\dfrac{{f(x)}}{{g(x)}}} \right)^1} = \dfrac{{f'(x)g(x) - f(x)g'(x)}}{{{{\left( {g(x)} \right)}^2}}}\] . Where \[f(x) = \sin (2x)\] and \[g(x) = \cos (2x)\] .
So we have,
 \[ = \dfrac{{\left( {\dfrac{d}{{dx}}\left( {\sin (2x)} \right) \times \cos (2x) - \sin (2x) \times \dfrac{d}{{dx}}\left( {\cos (2x)} \right)} \right)}}{{{{\left( {\cos (2x)} \right)}^2}}}\] .
We know \[\dfrac{d}{{dx}}\left( {\sin (2x)} \right) = 2\cos (2x)\] and \[\dfrac{d}{{dx}}\left( {\cos (2x)} \right) = - 2\sin (2x)\] , substituting we have,
 \[ = \dfrac{{\left( {2\cos (2x) \times \cos (2x) - \sin (2x) \times \left( { - 2\sin (2x)} \right)} \right)}}{{{{\left( {\cos (2x)} \right)}^2}}}\]
We know the product of two negative number will give us the positive number, we have
 \[ = \dfrac{{2{{\cos }^2}(2x) + 2{{\sin }^2}(2x)}}{{{{\cos }^2}(2x)}}\] .
Taking 2 common the numerator we have,
 \[ = \dfrac{{2\left( {{{\cos }^2}(2x) + {{\sin }^2}(2x)} \right)}}{{{{\cos }^2}(2x)}}\] .
We know the Pythagoras trigonometric identity. That Is \[{\cos ^2}(2x) + {\sin ^2}(2x) = 1\] .
 \[ = \dfrac{2}{{{{\cos }^2}(2x)}}\]
We know that the reciprocal of cosine is secant, we get
 \[ = 2{\sec ^2}(2x)\] .
Thus we have \[\dfrac{{dy}}{{dx}} = 2{\sec ^2}(2x)\] .
So, the correct answer is “$2{\sec ^2}(2x)$”.

Note: We have different types of differentiation rules.
Sum or difference rule: when the function is the sum or difference of two functions, the derivative is the sum or difference of derivative of each function. i.e.,
 \[f(x) = u(x) \pm v(x)\] then \[f'(x) = u'(x) \pm v'(x)\] .
Product rule: when f(x) is the product of two function that is \[f(x) = u(x).v(x)\] then \[f'(x) = u'(x).v(x) + u(x).v'(x)\] . We use this rule depending on the given problem.
In the problem of differentiation involving trigonometric function, to simplify further we need to know all the trigonometric identities. We know the reciprocal of sine, cosine and tangent are cosecant, secant and cotangent respectively.