Find the derivative using first principles? : # e^sinx#

Question

Ezra Adams · Accepted Answer

This is hardly from first principles but it is interesting.....

Start with the definition #e^z = 1 + z + z^2/(2!) + z^3/(3!)...# We can see that: #e^(sin x) = 1 + sin x + 1/(2!) sin^2 x + 1/(3!) sin^3 x ...# And so: #d/dx( e^(sin x) )= 0 + cos x + 2/(2!) sin x cos x + 3/(3!) sin^2 x cos x + ...# #= cos x( 1 + sin x + 1/(2!) sin^2 x + ...)# #= cos x * e^(sin x)#

Ezra Adams · Answer

#d/dx( e^sinx) = e^sinx cosx#

The definition of the derivative of #y=f(x)# is

# f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h #

So Let # f(x) = e^sinx # then the derivative of #y=f(x)# is given by:

# f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h # # " " = lim_(h rarr 0) ( e^sin(x+h)-e^sinx )/h # # " " = lim_(h rarr 0) ( e^sinx ( e^(sin(x+h)-sinx) - 1 ) ) / h # # " "= e^sinx lim_(h rarr 0) ( e^(sin(x+h)-sinx) - 1 ) / h #

# " " = e^sinx lim_(h rarr 0) ( e^(sin(x+h)-sinx) - 1 ) / h * (sin(x+h)-sinx)/(sin(x+h)-sinx) #

# " " = e^sinx lim_(h rarr 0) ( e^(sin(x+h)-sinx) - 1 ) / (sin(x+h)-sinx) * (sin(x+h)-sinx)/h #

# " " = e^sinx * L_1 * L_2#

Where:

# L_1 = lim_(h rarr 0) ( e^(sin(x+h)-sinx) - 1 ) / (sin(x+h)-sinx)#

# L_2 = lim_(h rarr 0) (sin(x+h)-sinx)/h#

Let us examine the first limit, #L_1#. Let # alpha=(sin(x+h)-sinx)# then #alpha rarr 0# as # h rarr 0# and so

# L_1 = lim_(h rarr 0) ( e^(sin(x+h)-sinx) - 1 ) / (sin(x+h)-sinx) # # \ \ \ \ = lim_(alpha rarr 0) ( e^alpha - 1 ) / alpha #

Now # lim_(alpha rarr 0) ( e^alpha - 1 ) / alpha = 1# is a standard calculus limit and so

# L_1 = 1#

Next we examine the second limit, #L_2#, We can use the sum and product formula:

#sin A - sin B = 2 cos((A + B)/2) sin ((A - B)/2) #

And we get:

# L_2 = lim_(h rarr 0) (sin(x+h)-sinx)/h # # \ \ \ \ = lim_(h rarr 0) (2cos((x+h+x)/2) sin ((x+h-x)/2))/h # # \ \ \ \ = lim_(h rarr 0) (2cos((2x+h)/2) sin (h/2))/h # # \ \ \ \ = lim_(h rarr 0) (cos(x+h/2) sin (h/2))/(h/2) # # \ \ \ \ = lim_(h rarr 0) cos(x+h/2) * lim_(h rarr 0) (sin (h/2))/(h/2) #

Let # beta = h/2# then #beta rarr 0# as #h rarr 0#, so:

# L_2 = lim_(h rarr 0) cos(x+h/2) * lim_(beta rarr 0) (sin (beta))/(beta)#

And #lim_(beta rarr 0) (sin (beta))/(beta) =1# is another standard calculus limit, giving us:

# L_2 = lim_(h rarr 0) cos(x+h/2) * 1# # \ \ \ \ = cos(x) #

Combining our results for #L_1# and #L_2# with our earlier result gives us:

# f'(x)=e^sinx * L_1 * L_2# # " "=e^sinx * 1 * cosx# # " "=e^sinx cosx#

Paisley Johnson · Answer

undefined To find the derivative of $ e^{\sin(x)} $ using first principles, we'll use the definition of the derivative:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substitute $ f(x) = e^{\sin(x)} $:

$$ f'(x) = \lim_{h \to 0} \frac{e^{\sin(x + h)} - e^{\sin(x)}}{h} $$

Apply the property of exponentials:

$$ f'(x) = \lim_{h \to 0} \frac{e^{\sin(x)} \cdot e^{\sin(h)} - e^{\sin(x)}}{h} $$

Factor out $ e^{\sin(x)} $:

$$ f'(x) = e^{\sin(x)} \lim_{h \to 0} \frac{e^{\sin(h)} - 1}{h} $$

As $ h \to 0 $, $ \sin(h) $ approaches 0, and $ e^{\sin(h)} $ approaches 1:

$$ f'(x) = e^{\sin(x)} \cdot 1 $$

Therefore, the derivative of $ e^{\sin(x)} $ with respect to $ x $ is $ e^{\sin(x)} $.