Explanation:
Step 1. We are given the expressions for u(x) and v(x):
[tex]\begin{gathered} u(x)=sin(x) \\ v(x)=x^{14} \end{gathered}[/tex]The first two parts of the problem consist of finding the derivative of these two expressions: u'(x) and v'(x).
Step 2. To derivate u(x) we use the following rule:
[tex]\begin{gathered} for\text{ a function } \\ g(x)=sin(x) \\ The\text{ derivative is} \\ g^{\prime}(x)=cos(x) \end{gathered}[/tex]which means that in this case:
[tex]\boxed{u^{\prime}(x)=cos(x)}[/tex]Step 3. To derivate v(x) we use the following rule:
[tex]\begin{gathered} for\text{ a function} \\ g(x)=x^n \\ The\text{ derivative is} \\ g^{\prime}(x)=nx^{n-1} \end{gathered}[/tex]In our case n=14, therefore, the derivative is:
[tex]\begin{gathered} v(x)=x^{14} \\ \downarrow \\ v^{\prime}(x)=14x^{14-1} \end{gathered}[/tex]simplifying the exponent:
[tex]\boxed{v^{\prime}(x)=14x^{13}}[/tex]Step 4. Now, to solve the third part of the problem, we consider the definition of f(x) given in the statement:
[tex]f(x)=\frac{u(x)}{v(x)}[/tex]And to find the derivative of this function f'(x) or f', we use the quotient rule,
[tex]f^{\prime}=\frac{u^{\prime}v-uv^{\prime}}{v^2}[/tex]We already know u and v from the given definitions, and we found u' and v' in 2 and 3.
So now, we substitute the known values into the quotient rule formula:
[tex]f^{\prime}=\frac{cos(x)(x^{14})-sin(x)(14x^{13})}{(x^{14})^2}[/tex]Step 5. The last step is to simplify our result. We start by simplifying the exponent in the denominator:
[tex]f^{\prime}=\frac{cos(x)(x^{14})-s\imaginaryI n(x)(14x^{13})}{x^{28}}[/tex]and to simplify further, divide both the numerator and denominator by x^13
[tex]\begin{gathered} f^{\prime}=\frac{cos(x)(x^)-s\imaginaryI n(x)(14)}{x^{15}} \\ \downarrow \\ \boxed{f^{\prime}=\frac{xcos(x)-14sin(x)}{x^{15}}} \end{gathered}[/tex]And that is the simplified solution.
Answer:
[tex]\begin{gathered} u^{\prime}(x)=cos(x) \\ v^{\prime}(x)=14x^{13} \\ f^{\prime}=\frac{xcos(x)-14s\imaginaryI n(x)}{x^{15}} \end{gathered}[/tex]