Differentiate e^cos x. The derivative of e^{cos x} is -sin x e^{cos x}. In this post, we will calculate the derivative of e to the power cos x by the first principle of derivatives as well as by the logarithmic differentiation and the chain rule of derivatives. We will also solve a few questions related to the differentiation of e^{cos x}.

Table of Contents

**What is the derivative of e**^{cos x}?

^{cos x}?

To find the derivative of e to the power cos x, we will first apply the chain rule of derivatives. The chain rule says that if f is a function of u, then the derivative of f with respect to x is

$\dfrac{df}{dx}=\dfrac{df}{du} \cdot \dfrac{du}{dx}$.

Step 1: Note that e^{cos x} is a function of cos x. So here f(u) = e^{u} with u=cos x.

Step 2: As u = cos x, we have

$\dfrac{du}{dx}=-\sin x$

Step 3: Now, by the above chain rule of derivatives, the derivative of e^{cos x} is

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

$=\dfrac{d}{du}\left(e^u \right) \cdot \dfrac{du}{dx}$

$=e^u \cdot (-\sin x)$ (putting the value of du/dx from above)

$=-\sin x e^{\cos x}$ as u=cos x.

So the derivative of e^{cos x} by the chain rule is -sin x e^{sin x}. In other words, d/dx(e^{cos x})=-sin x e^{ cos x}.

**Also Read:**

Derivative of e^{sin x}: The derivative of e^{sin x }is cos x e^{sin x}.Derivative of square root of x: The derivative of root x is 1/2√x. |

**Derivative of e**^{cos x} from first principle

^{cos x}from first principle

Now we will find the derivative of e^{cos x} using the first principle of derivatives. The derivative of a function f(x) by first principle is given by the following limit:

$\dfrac{d}{dx}\left( f(x) \right)$ $= \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Let f(x)=e^{cos x}. So the derivative of e^{cos x} by first principle is

$\dfrac{d}{dx}\left( e^{\cos x} \right)$ $= \lim\limits_{h \to 0} \dfrac{e^{\cos(x+h)} – e^{\cos x}}{h}$

$=\lim\limits_{h \to 0} \dfrac{e^{\cos x}(e^{\cos(x+h)-\cos x} -1)}{h}$

$=e^{\cos x}\lim\limits_{h \to 0} \dfrac{e^{\cos(x+h)-\cos x} -1}{h}$

$=e^{\cos x}\lim\limits_{h \to 0} \dfrac{e^{\cos(x+h)-\cos x} -1}{\cos(x+h)-\cos x}$ $\times \lim\limits_{h \to 0} \dfrac{\cos(x+h)-\cos x}{h}$

[Let t=cos(x+h)-cos x. Then t→0 as h→0]

$=e^{\cos x} \lim\limits_{t \to 0} \dfrac{e^t-1}{t}$ $\times \lim\limits_{h \to 0} \dfrac{-2\sin(x+h/2)\sin h/2}{h}$ as we know that cos c – cos d= -2 sin(c+d)/2 sin(c-d)/2.

$=-e^{\cos x} \times 1$ $\times \lim\limits_{h \to 0}\sin(x+h/2)$ $\times \lim\limits_{h \to 0} \dfrac{\sin h/2}{h/2}$ as the limit of (e^{t}-1)/t is 1 when t tends to 0.

$=-e^{\cos x} \sin(x+0/2)$ $\times \lim\limits_{z \to 0} \dfrac{\sin z}{z}$ where z=h/2.

$=-e^{\cos x} \sin x \times 1$ as $\lim\limits_{x \to 0} \dfrac{\sin x}{x}=1$

$=-\sin x e^{\cos x}$

Hence the derivative of e^{cos x} is equal to -sin x e^{cos x} , obtained by the first principle of derivatives.

**Also Read:**

Derivative of e^{2x}: The derivative of e^{2x }is 2e^{2x}.Derivative of cube root of x: The derivative of cube root x is (1/3)x ^{-2/3}. |

**Derivative of e**^{cos x} by substitution

^{cos x}by substitution

Next, we evaluate the derivative of e to the power cos x by the substitution method. The logarithmic derivatives will be used here.

Step 1: Let u=e^{cos x}. We need to find du/dx.

Step 2: Taking logarithm on both sides, we get

log_{e}u = log_{e}e^{cos x}

⇒ log_{e}u = cos x log_{e}e

⇒ log_{e}u = cos x as we know that log_{a}a=1.

Step 3: Differentiating with respect to x, we get that

$\dfrac{d}{dx}\left(\log_e u \right)=\dfrac{d}{dx}\left(\cos x \right)$

⇒ $\dfrac{1}{u} \dfrac{du}{dx}=-\sin x$

⇒ $\dfrac{du}{dx}=-u \sin x =-e^{\cos x} \sin x$ as u=e^{cos x}

Therefore, we obtain the derivative of e to the power cos x by the logarithmic differentiation which is equal to -sin x e^{cos x}.

**Question Answer on Derivative of e**^{cos x}

^{cos x}

**Question 1:** What is the derivative of xe^{cos x}?

*Answer:*

By product rule, the derivative of xe^{cos x} is

= x d/dx(e^{cos x}) + e^{cos x} d/dx(x)

= x (-sin xe^{cos x}) + e^{cos x} $\cdot$ 1

= -x sin xe^{cos x} + e^{cos x}

= e^{cos x} (-x sin x+1)

**Question 2:** Find the second derivative of e^{cos x}?

*Answer:*

As the first derivative of e^{cos x} is -sin x e^{cos x}, the second derivative of e to the power cos x is equal to

= d/dx(-sin x e^{cos x})

= -[sin x d/dx(e^{cos x})+ e^{cos x }d/dx(sin x)] by the product rule of derivatives.

= -[sin x (-sin x e^{cos x})+e^{cos x}(-cos x)]

= (sin^{2}x-cos x)e^{cos x}

So the second order derivative of e to the power cos x is (sin^{2}x-cos x)e^{cos x}.

**FAQs on Derivative of e**^{cos x}

^{cos x}

**Q1: What is the derivative of**

**e**?^{cos x}Answer: The derivative of e^{cos x} is e^{cos x} × d/dx(cos x) = -sin x e^{cos x}. Here we have applied the chain rule of derivatives.

**Q2: What is the derivative of**

**e**?^{sin x}Answer: The derivative of e^{sin x} is e^{sin x} × d/dx(sin x) = cos x e^{sin x}.

**Q3: What is the derivative of**

**e**?^{cos (x^2)}Answer: The derivative of e^{cos (x^2)} is e^{cos (x^2)} × d/dx(cos x^2) = -sin(x^2) e^{cos(x^2)} d/dx(x^2) = -2x sin(x^2) e^{cos(x^2)}. Here we have used the fact that the derivative of cos x is -sin x and the derivative of x^{2} is 2x.