Derivative of e^sinx: Formula, Proof by First Principle, Chain Rule

The derivative of e^sinx is equal to cosx e^sinx. Here we will calculate the derivative of e to the power sin x by the first principle, using the substitution method and the chain rule of derivatives. As an application, we will find the differentiation of several functions involving e^{sin x}.

What is the derivative of e^{sin x}?

Let us find the derivative of e^{sin x} by the chain rule of derivatives. Recall the chain rule: $\dfrac{df}{dx}=\dfrac{df}{du} \cdot \dfrac{du}{dx}$, where f is a function of u.

Step 1: Put u=sin x

Step 2: Differentiating with respect to x, we get

$\dfrac{du}{dx}$ = cos x

Step 3: Now, $\dfrac{d}{dx}\left(e^{\sin x} \right)=\dfrac{d}{dx}\left(e^u \right)$

$=\dfrac{d}{du}\left(e^u \right) \cdot \dfrac{du}{dx} \quad$ by the chain rule

= e^u cosx

= cosx e^sinx as u=sin x.

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

Also Read:

Integration of mod x : The integration of |x| is -x|x|/2+c Derivative of loge 3x : The derivative of loge 3x is 1/x. Derivative of sin 3x : The derivative of sin 3x is 3cos 3x.

Derivative of e^{sin x} from first principle

Now we will find the derivative of e^{sin x} using the limit definition of derivatives. Let f(x)=e^{sin x}. Then the derivative of f(x) by first principle is given as follows:

$\dfrac{d}{dx}\left( f(x) \right)$ = lim_h→0 $\dfrac{f(x+h)-f(x)}{h}$

So $\dfrac{d}{dx}\left( e^{\sin x} \right)$ = lim_h→0 $\dfrac{e^{\sin(x+h)} – e^{\sin x}}{h}$

= lim_h→0 $\dfrac{e^{\sin x}(e^{\sin(x+h)-\sin x} -1)}{h}$

= e^sinx lim_h→0 $\dfrac{e^{\sin(x+h)-\sin x} -1}{h}$

= e^sinx lim_h→0 $\dfrac{e^{\sin(x+h)-\sin x} -1}{\sin(x+h)-\sin x}$ × lim_h→0 $\dfrac{\sin(x+h)-\sin x}{h}$

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

= e^sinx lim_t→0 $\dfrac{e^t-1}{t}$ × lim_h→0 $\dfrac{2\cos(x+h/2)\sin h/2}{h}$ as we know that sin c – sin d= 2 cos(c+d)/2 sin(c-d)/2.

= e^sinx × 1 × lim_h→0 $\cos(x+h/2)$ × lim_h→0 $\dfrac{\sin h/2}{h/2}$

= e^sinx cos(x+0/2) × lim_z→0 sinz/z where z=h/2.

= e^sinx cosx × 1 as lim_x→0 sinx/x=1

= cosx e^sinx

So the derivative of e^sinx (e to the power sin x) from first principle is equal to e^sinx cosx.

Also Read:

Derivative of ecos x: The derivative of ecos x is -sin x ecos x. Derivative of square root of x: The derivative of root x is 1/2√x.

Derivative of e^{sin x} by substitution method

Next, we evaluate the derivative of e to the power sin x by the substitution method. Here we will use the logarithmic derivatives.

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

Step 2: Taking logarithm on both sides, we get

log_eu = log_ee^{sin x}

⇒ log_eu = sin x log_ee

⇒ log_eu = sin x as we know that log_aa=1.

Step 3: Differentiating we get that

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

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

⇒ $\dfrac{du}{dx}$ = u cosx = e^sinx cosx as u=e^{sin x}

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

Application of the derivative of e^{sin x}

As an application, we will find the differentiation of e^{sin 2x}. Let f(x)=e^{sin x}. So we have f(2x)=e^{sin 2x}.

From above we have: $\dfrac{d}{dx}\left( f(x) \right)$ = cos x e^{sin x}.

Note that we have to find the derivative of f(2x). So we put z=2x and hence dz/dx=2.

Now the derivative of f(2x) is equal to

$\dfrac{d}{dx}\left( f(2x) \right)=\dfrac{d}{dx}\left( f(z) \right)$

$=\dfrac{d}{dz}\left( f(z) \right) \cdot \dfrac{dz}{dx} \quad$ by the chain rule

= cosz e^sinz × 2

= 2cos2x e^sin2x as z=2x.

Hence the derivative of e^sin2x is 2cos2x e^sin2x.

Also Read:

Derivative of xlogx: The derivative of xlogx is 1+logx. Derivative of xex: The derivative of root xex is (1+x)ex.

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