Of course! Let's find the derivative of the function [tex]\( y = \left(x^{1/7} - 35\right) x^{-2} \)[/tex].
First, let's clarify the function we need to differentiate:
[tex]\[ y = (x^{1/7} - 35) x^{-2} \][/tex]
To differentiate this product, we will use the product rule, which states:
[tex]\[ \frac{d}{dx} [u \cdot v] = u' \cdot v + u \cdot v' \][/tex]
In our case, we will set:
[tex]\[ u = x^{1/7} - 35 \][/tex]
[tex]\[ v = x^{-2} \][/tex]
Now, we'll find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
1. Differentiate [tex]\( u = x^{1/7} - 35 \)[/tex]:
[tex]\[ u' = \frac{d}{dx} (x^{1/7} - 35) = \frac{1}{7} x^{1/7 - 1} \][/tex]
[tex]\[ u' = \frac{1}{7} x^{-6/7} \][/tex]
2. Differentiate [tex]\( v = x^{-2} \)[/tex]:
[tex]\[ v' = \frac{d}{dx} (x^{-2}) = -2 x^{-3} \][/tex]
Now we apply the product rule:
[tex]\[ \frac{dy}{dx} = u' \cdot v + u \cdot v' \][/tex]
Substituting [tex]\( u, u', v, \)[/tex] and [tex]\( v' \)[/tex] into the product rule:
[tex]\[ \frac{dy}{dx} = \left(\frac{1}{7} x^{-6/7}\right) \cdot x^{-2} + (x^{1/7} - 35) \cdot (-2 x^{-3}) \][/tex]
We now simplify each term:
1. Simplify [tex]\( \left(\frac{1}{7} x^{-6/7}\right) \cdot x^{-2} \)[/tex]:
[tex]\[ \left(\frac{1}{7} x^{-6/7}\right) \cdot x^{-2} = \frac{1}{7} x^{-6/7 - 2} = \frac{1}{7} x^{-6/7 - 14/7} = \frac{1}{7} x^{-20/7} \][/tex]
2. Simplify [tex]\((x^{1/7} - 35) \cdot (-2 x^{-3}) \)[/tex]:
[tex]\[ (x^{1/7} - 35) \cdot (-2 x^{-3}) = -2 (x^{1/7 - 3}) + 70 x^{-3} \][/tex]
[tex]\[ -2 (x^{1/7 - 3}) + 70 x^{-3} = -2 x^{-20/7} + 70 x^{-3} \][/tex]
Combining these simplified terms:
[tex]\[ \frac{dy}{dx} = \frac{1}{7} x^{-20/7} - 2 x^{-20/7} + 70 x^{-3} \][/tex]
Factorizing the common terms:
[tex]\[ \frac{dy}{dx} = \left(\frac{1}{7} - 2\right) x^{-20/7} + 70 x^{-3} \][/tex]
[tex]\[ \frac{dy}{dx} = \left(-\frac{13}{7}\right) x^{-20/7} + 70 x^{-3} \][/tex]
So, the derivative of the function [tex]\( y = (x^{1/7} - 35) x^{-2} \)[/tex] is:
[tex]\[ \frac{dy}{dx} = 0.142857 \frac{1}{x^{2.85714}} - \frac{2(x^{0.142857} - 35)}{x^{3}} \][/tex]
This gives us our final answer.
