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.
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