Respuesta :
Given the function:
[tex]f(x)=\sin (3x)[/tex]Let's find the function after differentiating it 68 times.
To differentiate, let's take the derivative.
First derivative:
[tex]\begin{gathered} f\text{'(x)= cos(}3x)\frac{d}{dx}(3x)_{} \\ \\ f^{\prime}(x)=3\cos (3x) \end{gathered}[/tex]Second derivative:
Since 3 is the constant with respect to x, the derivative of 3cos(3x) with respect to x will be:
[tex]\begin{gathered} f^{\doubleprime}(x)=3\frac{d}{dx}(\cos (3x)) \\ \\ f^{\doubleprime}(x)=3(-\sin (3x)\frac{d}{dx}(3x)) \\ \\ f^{\doubleprime}(x)=-3\sin (3x)(3\frac{d}{dx}(x)) \\ \\ f^{\doubleprime}(x)=-9\sin (3x)\frac{d}{dx}(x) \\ \\ f^{\doubleprime}(x)=-3^2\sin (3x) \end{gathered}[/tex]After the second differentiation, we have -sin, , the same will be applicable to the 68th time.
Therefore, for the 68th differentiation the exponent for the constant -3 will be 68.
[tex]f(x)=-3^{68}\sin (3x)[/tex]Therefore, after differentiating the function 68 times, we have:
[tex]f(x)=-3^{68}\sin (3x)[/tex]ANSWER:
[tex]f(x)=-3^{68}\sin (3x)[/tex]