How do you find f'(x) using the definition of a derivative #y=6e^x+4/root3x#?

Question

Luke Alvarez · Accepted Answer

#6e^x-4/(3x^(4/3#

#\frac{d}{dx}(6e^x+\frac{4}{\sqrt[3]{x}})#

Applying sum/difference rule: #(f\pm g)^'=f^'\pm g^'#

#=\frac{d}{dx}(6e^x)+\frac{d}{dx}(\frac{4}{\sqrt[3]{x}})#

We have, #=\frac{d}{dx}(6e^x)# = #6e^x#

(Taking the constant out : #(a\cdot f)^'=a\cdot f^'# i.e #=6\frac{d}{dx}(e^x)# and applying common derivative of #\frac{d}{dx}(e^x)=e^x#)

And,

#\frac{d}{dx}(\frac{4}{\sqrt[3]{x}})=-\frac{4}{3x^{\frac{4}{3}}}#

(Taking the constant out : #(a\cdot f)^'=a\cdot f^'# i.e #=4\frac{d}{dx}(\frac{1}{\sqrt[3]{x}})# and applying power rule for #=4\frac{d}{dx}(x^{-\frac{1}{3}})# i.e #\frac{d}{dx}(x^a)=a\cdot x^{a-1}# we get #=-\frac{4}{3x^{\frac{4}{3}}}#)

So, finally #6e^x# - #\frac{4}{3x^{\frac{4}{3}}}#

Cameron Alexander · Answer

undefined To find $ f'(x) $ using the definition of a derivative for the function $ y = 6e^x + \frac{4}{\sqrt{3}x} $, we can use the formula:

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

First, we need to find $ f(x + h) $:
$$ f(x + h) = 6e^{x + h} + \frac{4}{\sqrt{3}(x + h)} $$

Next, we subtract $ f(x) $ from $ f(x + h) $:
$$ f(x + h) - f(x) = 6e^{x + h} + \frac{4}{\sqrt{3}(x + h)} - (6e^x + \frac{4}{\sqrt{3}x}) $$

Then, we simplify the expression:
$$ f(x + h) - f(x) = 6e^x e^h - 6e^x + \frac{4}{\sqrt{3}(x + h)} - \frac{4}{\sqrt{3}x} $$

Now, we divide the result by $ h $:
$$ \frac{f(x + h) - f(x)}{h} = \frac{6e^x e^h - 6e^x}{h} + \frac{4}{\sqrt{3}h(x + h)} - \frac{4}{\sqrt{3}hx} $$

Finally, we take the limit as $ h $ approaches 0 to find $ f'(x) $:
$$ f'(x) = \lim_{h \to 0} \left( \frac{6e^x e^h - 6e^x}{h} + \frac{4}{\sqrt{3}h(x + h)} - \frac{4}{\sqrt{3}hx} \right) $$

After evaluating the limit, we obtain the derivative $ f'(x) $.