How do you find the derivative by definition for #y=x^(7/3)#?

Question

Scarlett Allison · Accepted Answer

The details depend on whether you use  #f'(a) = lim_(xrarra) (x^(7/3) - a^(7/3))/(x-a)# or, #f'(a) = lim_(hrarr0) ( (a+h)^(7/3) - a^(7/3))/(h)#. Either way, we'll start with the same algebra: #u^3 - v^3 = (u-v) (u^2 +uv + v^2)#. So we can 'rationalize' differences involving cube roots: #u - v  = (u^(1/3) -v^(1/3) ) (u^(2/3)  +u^(1/3) v^(1/3)  + v^(2/3) )#. This may be easier to see written: #u - v  = (root(3)u-root(3)v ) (root(3)u^2  +root(3)(u v)  + root(3)v^2 )#. This allows us to rewrite the fraction with a numerator that is a difference of 7th powers. Then, depending on the form of the definition you are using: Factor:
#x^7 - a^7 = (x-a)(x^6+x^5a + x^4a^2 + x^3a^3 + x^2a^4 + xa^5 + a^6#. Or, expand (using the binomial theorem) #(a+h)^7 - a^7 = [a^7 + 7a^6h + ((7),(2))a^5h^2 + ((7),(3))a^4h^3 + * * *  + 7ah^6 + h^7] - a^7 #. Finally, reduce the fraction and evaluate the limit.

Adam Adkins · Answer

undefined To find the derivative of $ y = x^{\frac{7}{3}} $ by definition, we use the definition of the derivative:

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

Substitute $ f(x) = x^{\frac{7}{3}} $ into the definition:

$$ f'(x) = \lim_{h \to 0} \frac{(x+h)^{\frac{7}{3}} - x^{\frac{7}{3}}}{h} $$

Expand $ (x+h)^{\frac{7}{3}} $ using binomial theorem:

$$ f'(x) = \lim_{h \to 0} \frac{(x^{\frac{7}{3}} + \frac{7}{3}x^{\frac{4}{3}}h + \frac{21}{9}x^{\frac{1}{3}}h^2 + \frac{35}{27}h^3) - x^{\frac{7}{3}}}{h} $$

Simplify the expression:

$$ f'(x) = \lim_{h \to 0} \frac{\frac{7}{3}x^{\frac{4}{3}}h + \frac{21}{9}x^{\frac{1}{3}}h^2 + \frac{35}{27}h^3}{h} $$

$$ f'(x) = \lim_{h \to 0} \left( \frac{7}{3}x^{\frac{4}{3}} + \frac{21}{9}x^{\frac{1}{3}}h + \frac{35}{27}h^2 \right) $$

Now, as $ h \to 0 $, the term $ \frac{21}{9}x^{\frac{1}{3}}h $ and $ \frac{35}{27}h^2 $ approach 0:

$$ f'(x) = \frac{7}{3}x^{\frac{4}{3}} $$

Therefore, the derivative of $ y = x^{\frac{7}{3}} $ by definition is $ \frac{7}{3}x^{\frac{4}{3}} $.