
Find the derivative of the following function:
${\text{cosec }}x{\text{ }}\cot x$
Answer
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Hint: In this question apply the product rule of differentiation which is given as $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ later on in the solution apply the differentiation property of cosec x and cos x which is given as $\dfrac{d}{{dx}}\left( {{\text{cosec }}x} \right) = - {\text{cosec }}x\cot x{\text{ and }}\dfrac{d}{{dx}}\left( {\cot x} \right) = - {\text{cose}}{{\text{c}}^2}x$ so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Let
$y = {\text{cosec }}x{\text{ }}\cot x$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {{\text{cosec }}x{\text{ }}\cot x} \right]$
Now here we use product rule of differentiate which is given as
$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {{\text{cosec }}x} \right)\dfrac{d}{{dx}}\left( {\cot x} \right) + \left( {\cot x} \right)\dfrac{d}{{dx}}\left( {{\text{cosec }}x} \right)$
Now as we know that differentiation of $\dfrac{d}{{dx}}\left( {{\text{cosec }}x} \right) = - {\text{cosec }}x\cot x{\text{ and }}\dfrac{d}{{dx}}\left( {\cot x} \right) = - {\text{cose}}{{\text{c}}^2}x$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {{\text{cosec }}x} \right)\left( { - {\text{cose}}{{\text{c}}^2}x} \right) + \left( {\cot x} \right)\left( { - \cos ecx\cot x} \right)$
Now simplify it we have,
$ \Rightarrow \dfrac{d}{{dx}}y = - \cos ecx\left[ {\cos e{c^2}x + {{\cot }^2}x} \right]$
So this is the required differentiation.
Note – Whenever we face such types of questions the key concept is always recall the formula of product rule of differentiation, formula of cosec x and cot x differentiation which is stated above then first apply the product rule as above then use the property of differentiation of cosec x and cot x as above and simplify we will get the required answer.
Complete step-by-step answer:
Let
$y = {\text{cosec }}x{\text{ }}\cot x$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {{\text{cosec }}x{\text{ }}\cot x} \right]$
Now here we use product rule of differentiate which is given as
$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {{\text{cosec }}x} \right)\dfrac{d}{{dx}}\left( {\cot x} \right) + \left( {\cot x} \right)\dfrac{d}{{dx}}\left( {{\text{cosec }}x} \right)$
Now as we know that differentiation of $\dfrac{d}{{dx}}\left( {{\text{cosec }}x} \right) = - {\text{cosec }}x\cot x{\text{ and }}\dfrac{d}{{dx}}\left( {\cot x} \right) = - {\text{cose}}{{\text{c}}^2}x$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {{\text{cosec }}x} \right)\left( { - {\text{cose}}{{\text{c}}^2}x} \right) + \left( {\cot x} \right)\left( { - \cos ecx\cot x} \right)$
Now simplify it we have,
$ \Rightarrow \dfrac{d}{{dx}}y = - \cos ecx\left[ {\cos e{c^2}x + {{\cot }^2}x} \right]$
So this is the required differentiation.
Note – Whenever we face such types of questions the key concept is always recall the formula of product rule of differentiation, formula of cosec x and cot x differentiation which is stated above then first apply the product rule as above then use the property of differentiation of cosec x and cot x as above and simplify we will get the required answer.
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