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Question: How do you find the derivative of \[\left( {\cos e{c^2}x} \right)\] ?...

How do you find the derivative of (cosec2x)\left( {\cos e{c^2}x} \right) ?

Explanation

Solution

In the given problem, we are required to differentiate (cosec2x)\left( {\cos e{c^2}x} \right) with respect to x. Since, (cosec2x)\left( {\cos e{c^2}x} \right) is a composite function, so we will have to apply chain rule of differentiation in the process of differentiating (cosec2x)\left( {\cos e{c^2}x} \right) . So, differentiation of (cosec2x)\left( {\cos e{c^2}x} \right) with respect to x will be done layer by layer using the chain rule of differentiation. The derivative of cosec(x)\cos ec\left( x \right)with respect to x must be remembered.

Complete step by step answer:
To find derivative of (cosec2x)\left( {\cos e{c^2}x} \right) with respect to xx, we have to find differentiate y=(cosec2x)y = \left( {\cos e{c^2}x} \right)with respect to xx. So, Derivative of y=(cosec2x)y = \left( {\cos e{c^2}x} \right) with respect to xxcan be calculated as dydx=ddx(cosec2x)\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\cos e{c^2}x} \right) .
Now, dydx=ddx(cosec2x)\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\cos e{c^2}x} \right)
Now, Let us assume u=cosec(x)u = \cos ec\left( x \right). So substituting cosec(x)\cos ec\left( x \right) as uu, we get,
\Rightarrow dydx=ddx[u2]\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {{u^2}} \right]
Now, we know the power rule of differentiation. So, according to the power rule of differentiation, the derivative of (xn)\left( {{x^n}} \right) is nxn1n{x^{n - 1}}. So, we get the derivative of u2{u^2} with respect to u as 2u2u by following the power rule of differentiation.

But, we will have to differentiate u by x again as it is also a variable. So, we get,
\Rightarrow dydx=(2u)dudx\dfrac{{dy}}{{dx}} = \left( {2u} \right)\dfrac{{du}}{{dx}}
Now, putting back uuas cosec(x)\cos ec\left( x \right), we get,
\Rightarrow dydx=(2cosec(x))ddx[cosecx]\dfrac{{dy}}{{dx}} = \left( {2\cos ec\left( x \right)} \right)\dfrac{d}{{dx}}\left[ {\cos ecx} \right]
Now, we know that the derivative of [cosecx]\left[ {\cos ecx} \right] with respect to x is [cosecx][cotx] - \left[ {\cos ecx} \right]\left[ {\cot x} \right]. Hence, we get,
\Rightarrow dydx=(2cosecx)[cosec(x)cot(x)]\dfrac{{dy}}{{dx}} = \left( {2\cos ecx} \right)\left[ { - \cos ec\left( x \right)\cot \left( x \right)} \right]
Simplifying further, we get,
\therefore dydx=2cosec2(x)cot(x)\dfrac{{dy}}{{dx}} = - 2\cos e{c^2}\left( x \right)\cot \left( x \right)

So, the derivative of (cosec2x)\left( {\cos e{c^2}x} \right) with respect to xx is [2cosec2(x)cot(x)]\left[ { - 2\cos e{c^2}\left( x \right)\cot \left( x \right)} \right].

Note: The given problem may also be solved using the first principle of differentiation. The derivatives of basic functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.