Differentiate e^cos x. The derivative of ecos x is -sin x ecos 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 ecos x.
Table of Contents
What is the derivative of ecos 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 ecos x is a function of cos x. So here f(u) = eu 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 ecos 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 ecos x by the chain rule is -sin x esin x. In other words, d/dx(ecos x)=-sin x e cos x.
Also Read:
| Derivative of esin x: The derivative of esin x is cos x esin x. Derivative of square root of x: The derivative of root x is 1/2√x. |
Derivative of ecos x from first principle
Now we will find the derivative of ecos 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)=ecos x. So the derivative of ecos 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 (et-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 ecos x is equal to -sin x ecos x , obtained by the first principle of derivatives.
Also Read:
| Derivative of e2x: The derivative of e2x is 2e2x. Derivative of cube root of x: The derivative of cube root x is (1/3)x-2/3. |
Derivative of ecos 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=ecos x. We need to find du/dx.
Step 2: Taking logarithm on both sides, we get
logeu = logeecos x
⇒ logeu = cos x logee
⇒ logeu = cos x as we know that logaa=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=ecos x
Therefore, we obtain the derivative of e to the power cos x by the logarithmic differentiation which is equal to -sin x ecos x.
Question Answer on Derivative of ecos x
Question 1: What is the derivative of xecos x?
Answer:
By product rule, the derivative of xecos x is
= x d/dx(ecos x) + ecos x d/dx(x)
= x (-sin xecos x) + ecos x $\cdot$ 1
= -x sin xecos x + ecos x
= ecos x (-x sin x+1)
Question 2: Find the second derivative of ecos x?
Answer:
As the first derivative of ecos x is -sin x ecos x, the second derivative of e to the power cos x is equal to
= d/dx(-sin x ecos x)
= -[sin x d/dx(ecos x)+ ecos x d/dx(sin x)] by the product rule of derivatives.
= -[sin x (-sin x ecos x)+ecos x(-cos x)]
= (sin2x-cos x)ecos x
So the second order derivative of e to the power cos x is (sin2x-cos x)ecos x.
FAQs on Derivative of ecos x
Answer: The derivative of ecos x is ecos x × d/dx(cos x) = -sin x ecos x. Here we have applied the chain rule of derivatives.
Answer: The derivative of esin x is esin x × d/dx(sin x) = cos x esin x.
Answer: The derivative of ecos (x^2) is ecos (x^2) × d/dx(cos x^2) = -sin(x^2) ecos(x^2) d/dx(x^2) = -2x sin(x^2) ecos(x^2). Here we have used the fact that the derivative of cos x is -sin x and the derivative of x2 is 2x.
This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.