Answer
396.9k+ views
Hint: In this problem, we find out the derivative using the chain rule of differentiation, which is $\dfrac{d}{dx}f\left( u\left( x \right) \right)=\dfrac{df}{du}\times \dfrac{du}{dx}$ . First, we differentiate ${{e}^{\cos x}}$ with respect to $\cos x$ and multiply the result with the derivative of $\cos x$ with respect to $x$ . By doing so, we arrive at the desired result.
Complete step by step solution:
The given equation we have is
$f\left( x \right)={{e}^{\cos x}}$
We can rewrite the above equation as
$\Rightarrow y={{e}^{\cos x}}$
Now for differentiation we apply chain rule for the right-hand part. According to the chain rule of differentiation: $\dfrac{d}{dx}f\left( u\left( x \right) \right)=\dfrac{df}{du}\times \dfrac{du}{dx}$
Here, the functions we have assumed are $f\left( u\left( x \right) \right)={{e}^{\cos x}}$ and $u\left( x \right)=\cos x$ .
Taking the main equation $y={{e}^{\cos x}}$ and differentiating both the sides, we get
$\dfrac{dy}{dx}=\dfrac{d\left\{ {{e}^{\cos x}} \right\}}{dx}$
Applying the chain rule of differentiation, we rewrite the above expression as,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{e}^{\cos x}} \right)}{d\left( \cos x \right)}\times \dfrac{d\left( \cos x \right)}{dx}$
Further carrying out the differentiation, we know that the derivative of exponential functions gives nothing different, but the function itself and that the derivative of $\cos x$ is $-\left( \sin x \right)$ . Implementing these in the above equation, the above equation thus becomes
$\Rightarrow \dfrac{dy}{dx}={{e}^{\cos x}}\times \left( -\sin x \right)$
Simplifying the above equation, the above equation thus becomes
$\Rightarrow \dfrac{dy}{dx}=-{{e}^{\cos x}}\sin x$
Therefore, we conclude that the derivative of the given equation $y={{e}^{\cos x}}$ is $-{{e}^{\cos x}}\sin x$ .
Note:
While applying the chain rule, we come across a lot of functions and a lot of derivatives. Thus, we are most likely to make mistakes here. So, we must be careful while dealing with the chain rule and must take care of the various along with the signs. The derivative of the given expression can also be found out by using the definition of differentiation. But, that method will become too tedious to carry out and thus it is not advisable to do so.
Complete step by step solution:
The given equation we have is
$f\left( x \right)={{e}^{\cos x}}$
We can rewrite the above equation as
$\Rightarrow y={{e}^{\cos x}}$
Now for differentiation we apply chain rule for the right-hand part. According to the chain rule of differentiation: $\dfrac{d}{dx}f\left( u\left( x \right) \right)=\dfrac{df}{du}\times \dfrac{du}{dx}$
Here, the functions we have assumed are $f\left( u\left( x \right) \right)={{e}^{\cos x}}$ and $u\left( x \right)=\cos x$ .
Taking the main equation $y={{e}^{\cos x}}$ and differentiating both the sides, we get
$\dfrac{dy}{dx}=\dfrac{d\left\{ {{e}^{\cos x}} \right\}}{dx}$
Applying the chain rule of differentiation, we rewrite the above expression as,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{e}^{\cos x}} \right)}{d\left( \cos x \right)}\times \dfrac{d\left( \cos x \right)}{dx}$
Further carrying out the differentiation, we know that the derivative of exponential functions gives nothing different, but the function itself and that the derivative of $\cos x$ is $-\left( \sin x \right)$ . Implementing these in the above equation, the above equation thus becomes
$\Rightarrow \dfrac{dy}{dx}={{e}^{\cos x}}\times \left( -\sin x \right)$
Simplifying the above equation, the above equation thus becomes
$\Rightarrow \dfrac{dy}{dx}=-{{e}^{\cos x}}\sin x$
Therefore, we conclude that the derivative of the given equation $y={{e}^{\cos x}}$ is $-{{e}^{\cos x}}\sin x$ .
Note:
While applying the chain rule, we come across a lot of functions and a lot of derivatives. Thus, we are most likely to make mistakes here. So, we must be careful while dealing with the chain rule and must take care of the various along with the signs. The derivative of the given expression can also be found out by using the definition of differentiation. But, that method will become too tedious to carry out and thus it is not advisable to do so.
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