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Hint- Here we can use the chain rule to solve this. To do that, we'll have to determine what the "outer" function is and what the "inner" function composed in the outer function is. After determining outer and inner function we can use the chain rule (F'(x)=f'(g(x)) (g'(x))) which is mainly used to differentiate the composite function.

In the given question, cot(x) is the "inner" function that is composed as part of the ${\text{cot }}{{\text{x}}^3}$..

The chain rule is:

F'(x)=f'(g(x)) (g'(x))

Language wise- the derivative of the outer function f(x) (with the inside function g(x) left alone!) times the derivative of the inner function.

In the given question- function f(x) = cot(x) and g(x)= ${{\text{x}}^3}$

(1) The derivative of the outer function = f(x) = cot(x) (with the inside function left alone) is:

Here we consider ${{\text{x}}^3}$ the same as x.

$\dfrac{{d\cot {\text{x}}}}{{dx}} = - {\text{cs}}{{\text{c}}^2}{\text{x}}$

2) The derivative of the inner function g(x)= ${{\text{x}}^3}$

$\dfrac{{d{{\text{x}}^3}}}{{dx}} = 3{{\text{x}}^2}$

Combining the two steps(1) and (2) through multiplication to get the derivative:

$\dfrac{{dy}}{{dx}} = \dfrac{{d\cot {{\text{x}}^3}}}{{dx}} = - {\text{cs}}{{\text{c}}^2}{{\text{x}}^3}.(3{{\text{x}}^2})$

\[\dfrac{{dy}}{{dx}} = - (3{{\text{x}}^2}){\text{cs}}{{\text{c}}^2}{{\text{x}}^3}.\]

Hence Differentiation of ${\text{y = cot }}{{\text{x}}^3}$ w.r.t x will be equal to $ - (3{{\text{x}}^2}){\text{cs}}{{\text{c}}^2}{{\text{x}}^3}$

Note- This particular problem can also be solved by letting ${{\text{x}}^3}$ equal to t.

Given: ${\text{y = cot }}{{\text{x}}^3}$

Put ${{\text{x}}^3}$=t

Differentiating w.r.t. x on both side

$3{{\text{x}}^2}$=$\dfrac{{dt}}{{dx}}$

${\text{y = cot }}{{\text{x}}^3} \Rightarrow \cot ({\text{t}})$

Differentiating w.r.t. x on both side

$\dfrac{{dy}}{{dx}}{\text{ = }}\dfrac{{d\cot ({\text{t}})}}{{dx}} \Rightarrow - {\text{cose}}{{\text{c}}^2}{\text{t}}\dfrac{{dt}}{{dx}}$

On putting ${{\text{x}}^3}$=t and, $3{{\text{x}}^2}$=$\dfrac{{dt}}{{dx}}$

$\dfrac{{dy}}{{dx}} = \dfrac{{d\cot {{\text{x}}^3}}}{{dx}} = - {\text{cs}}{{\text{c}}^2}{{\text{x}}^3}.(3{{\text{x}}^2})$

\[\dfrac{{dy}}{{dx}} = - (3{{\text{x}}^2}){\text{cs}}{{\text{c}}^2}{{\text{x}}^3}.\]

__Complete step-by-step solution -__In the given question, cot(x) is the "inner" function that is composed as part of the ${\text{cot }}{{\text{x}}^3}$..

The chain rule is:

F'(x)=f'(g(x)) (g'(x))

Language wise- the derivative of the outer function f(x) (with the inside function g(x) left alone!) times the derivative of the inner function.

In the given question- function f(x) = cot(x) and g(x)= ${{\text{x}}^3}$

(1) The derivative of the outer function = f(x) = cot(x) (with the inside function left alone) is:

Here we consider ${{\text{x}}^3}$ the same as x.

$\dfrac{{d\cot {\text{x}}}}{{dx}} = - {\text{cs}}{{\text{c}}^2}{\text{x}}$

2) The derivative of the inner function g(x)= ${{\text{x}}^3}$

$\dfrac{{d{{\text{x}}^3}}}{{dx}} = 3{{\text{x}}^2}$

Combining the two steps(1) and (2) through multiplication to get the derivative:

$\dfrac{{dy}}{{dx}} = \dfrac{{d\cot {{\text{x}}^3}}}{{dx}} = - {\text{cs}}{{\text{c}}^2}{{\text{x}}^3}.(3{{\text{x}}^2})$

\[\dfrac{{dy}}{{dx}} = - (3{{\text{x}}^2}){\text{cs}}{{\text{c}}^2}{{\text{x}}^3}.\]

Hence Differentiation of ${\text{y = cot }}{{\text{x}}^3}$ w.r.t x will be equal to $ - (3{{\text{x}}^2}){\text{cs}}{{\text{c}}^2}{{\text{x}}^3}$

Note- This particular problem can also be solved by letting ${{\text{x}}^3}$ equal to t.

Given: ${\text{y = cot }}{{\text{x}}^3}$

Put ${{\text{x}}^3}$=t

Differentiating w.r.t. x on both side

$3{{\text{x}}^2}$=$\dfrac{{dt}}{{dx}}$

${\text{y = cot }}{{\text{x}}^3} \Rightarrow \cot ({\text{t}})$

Differentiating w.r.t. x on both side

$\dfrac{{dy}}{{dx}}{\text{ = }}\dfrac{{d\cot ({\text{t}})}}{{dx}} \Rightarrow - {\text{cose}}{{\text{c}}^2}{\text{t}}\dfrac{{dt}}{{dx}}$

On putting ${{\text{x}}^3}$=t and, $3{{\text{x}}^2}$=$\dfrac{{dt}}{{dx}}$

$\dfrac{{dy}}{{dx}} = \dfrac{{d\cot {{\text{x}}^3}}}{{dx}} = - {\text{cs}}{{\text{c}}^2}{{\text{x}}^3}.(3{{\text{x}}^2})$

\[\dfrac{{dy}}{{dx}} = - (3{{\text{x}}^2}){\text{cs}}{{\text{c}}^2}{{\text{x}}^3}.\]

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