Differentiate the function. $f(t)=\frac{1}{2} t^{6}-3 t^{4}+t$

Solution

Verified

Step 1

1 of 2

\begin{gathered} f\left( t \right) = \frac{1}{2}{t^6} - 3{t^4} + t \\ \textcolor{#4257b2}{ {\text{Differentiate both sides}}} \\ f'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{1}{2}{t^6} - 3{t^4} + t} \right] \\ \textcolor{#4257b2}{ {\text{Apply the property }}\frac{d}{{dt}}\left[ {g\left( t \right) \pm h\left( t \right)} \right] = \frac{d}{{dt}}\left[ {g\left( t \right)} \right] \pm \frac{d}{{dt}}\left[ {h\left( t \right)} \right]} \\ f'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{1}{2}{t^6}} \right] - \frac{d}{{dt}}\left[ {3{t^4}} \right] + \frac{d}{{dt}}\left[ t \right] \\ f'\left( t \right) = \frac{1}{2}\frac{d}{{dt}}\left[ {{t^6}} \right] - 3\frac{d}{{dt}}\left[ {{t^4}} \right] + \frac{d}{{dt}}\left[ t \right] \\ \textcolor{#4257b2}{ {\text{Apply }}\frac{d}{{dt}}\left[ {{t^n}} \right] = n{t^{n - 1}}{\text{ and }}\frac{d}{{dt}}\left[ c \right] = 0,{\text{ so}}} \\ f'\left( t \right) = \frac{1}{2}\left( {6{t^5}} \right) - 3\left( {4{t^3}} \right) + \left( 1 \right) \\ \textcolor{#4257b2}{ {\text{Simplify}}} \\ f'\left( t \right) = 3{t^5} - 12{t^3} + 1 \\ \end{gathered}

Create a free account to view solutions

By signing up, you accept Quizlet's Terms of Service and Privacy Policy

By signing up, you accept Quizlet's Terms of Service and Privacy Policy