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Hint: Consider $A = {\sin ^2}x$ and $B = {e^{\cos x}}$, differentiate both of them with respect to x. We will be obtaining $\dfrac{{dA}}{{dx}}$ and$\dfrac{{dB}}{{dx}}$. Divide both of them to obtain the required $\dfrac{{dA}}{{dB}}$.

Complete step-by-step answer:

Let A = ${\sin ^2}x$ and B =${e^{\cos x}}$.

Now differentiate A and B w.r.t. x we have,

$ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{d}{{dx}}{\sin ^2}x$

Now differentiate according to property $\dfrac{d}{{dx}}{\sin ^m}x = m{\sin ^{m - 1}}x\left( {\dfrac{d}{{dx}}\sin x} \right)$

$ \Rightarrow \dfrac{{dA}}{{dx}} = 2\sin x\left( {\dfrac{d}{{dx}}\sin x} \right)$

Now as we know differentiate of sin x is cos x so use this we have,

$ \Rightarrow \dfrac{{dA}}{{dx}} = 2\sin x\left( {\cos x} \right)$ ........................... (2)

Now differentiate B w.r.t. x we have,

$ \Rightarrow \dfrac{{dB}}{{dx}} = \dfrac{d}{{dx}}{e^{\cos x}}$

Now differentiate above equation according to property $\dfrac{d}{{dx}}{e^{mx}} = {e^{mx}}\left( {\dfrac{d}{{dx}}mx} \right)$ we have,

$ \Rightarrow \dfrac{{dB}}{{dx}} = {e^{\cos x}}\left( {\dfrac{d}{{dx}}\cos x} \right)$

Now as we know differentiate of cos x is -sin x so use this we have,

$ \Rightarrow \dfrac{{dB}}{{dx}} = {e^{\cos x}}\left( { - \sin x} \right)$ ..................... (2)

Now divide equation (1) by equation (2) we have,

\[ \Rightarrow \dfrac{{\dfrac{{dA}}{{dx}}}}{{\dfrac{{dB}}{{dx}}}} = \dfrac{{2\sin x\left( {\cos x} \right)}}{{{e^{\cos x}}\left( { - \sin x} \right)}}\]

$ \Rightarrow \dfrac{{dA}}{{dB}} = \dfrac{{ - 2\cos x}}{{{e^{\cos x}}}}$

So this is the required differentiation of ${\sin ^2}x$ w.r.t. ${e^{\cos x}}$.

So this is the required answer.

Note: When the derivative of one quantity is to be found with respect to another quantity it is advised to look for the common variable in both the quantities and to derivative them with respect to that common entity. It is advised to remember derivatives of some common entities like sinx and ${e^x}$. The chain rule of derivatives helps finding generalized derivatives of such frequently used derivatives.

Complete step-by-step answer:

Let A = ${\sin ^2}x$ and B =${e^{\cos x}}$.

Now differentiate A and B w.r.t. x we have,

$ \Rightarrow \dfrac{{dA}}{{dx}} = \dfrac{d}{{dx}}{\sin ^2}x$

Now differentiate according to property $\dfrac{d}{{dx}}{\sin ^m}x = m{\sin ^{m - 1}}x\left( {\dfrac{d}{{dx}}\sin x} \right)$

$ \Rightarrow \dfrac{{dA}}{{dx}} = 2\sin x\left( {\dfrac{d}{{dx}}\sin x} \right)$

Now as we know differentiate of sin x is cos x so use this we have,

$ \Rightarrow \dfrac{{dA}}{{dx}} = 2\sin x\left( {\cos x} \right)$ ........................... (2)

Now differentiate B w.r.t. x we have,

$ \Rightarrow \dfrac{{dB}}{{dx}} = \dfrac{d}{{dx}}{e^{\cos x}}$

Now differentiate above equation according to property $\dfrac{d}{{dx}}{e^{mx}} = {e^{mx}}\left( {\dfrac{d}{{dx}}mx} \right)$ we have,

$ \Rightarrow \dfrac{{dB}}{{dx}} = {e^{\cos x}}\left( {\dfrac{d}{{dx}}\cos x} \right)$

Now as we know differentiate of cos x is -sin x so use this we have,

$ \Rightarrow \dfrac{{dB}}{{dx}} = {e^{\cos x}}\left( { - \sin x} \right)$ ..................... (2)

Now divide equation (1) by equation (2) we have,

\[ \Rightarrow \dfrac{{\dfrac{{dA}}{{dx}}}}{{\dfrac{{dB}}{{dx}}}} = \dfrac{{2\sin x\left( {\cos x} \right)}}{{{e^{\cos x}}\left( { - \sin x} \right)}}\]

$ \Rightarrow \dfrac{{dA}}{{dB}} = \dfrac{{ - 2\cos x}}{{{e^{\cos x}}}}$

So this is the required differentiation of ${\sin ^2}x$ w.r.t. ${e^{\cos x}}$.

So this is the required answer.

Note: When the derivative of one quantity is to be found with respect to another quantity it is advised to look for the common variable in both the quantities and to derivative them with respect to that common entity. It is advised to remember derivatives of some common entities like sinx and ${e^x}$. The chain rule of derivatives helps finding generalized derivatives of such frequently used derivatives.

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