# How do you differentiate f(t)=1/2t^6-3t^4+1?

Oct 30, 2016

We can differentiate term by term. (Things that are added together are called "terms".)

So we need to differentiate $\frac{1}{2} {t}^{6}$ and $- 3 {t}^{4}$ and $1$. (We don't need to write all of this, but I'm explaining our thought process.)

To differentiate $\frac{1}{2} {t}^{6}$ we will used the power rule. The constant $\frac{1}{2}$ just hangs out out front.
In more proper language: the derivative of $\frac{1}{2}$ times ${t}^{6}$ is $\frac{1}{2}$ times the derivative of ${t}^{6}$.

In notation: $\frac{d}{\mathrm{dt}} \left(\frac{1}{2} {t}^{6}\right) = \frac{1}{2} \frac{d}{\mathrm{dt}} \left({t}^{6}\right)$

Now, we find the derivative of $t \wedge 6$. Multiply by the exponent, then subtract one from the exponent to get the new exponent.

$6 {t}^{6 - 1} = 6 {t}^{5}$

We use the same process to differentiate $- 3 {t}^{4}$.

The derivative of a constant, like $1$, is $0$

The derivative of $f \left(t\right) = \frac{1}{2} {t}^{6} - 3 {t}^{4} + 1$ is

$f ' \left(t\right) = \frac{1}{2} \left(6 {t}^{5}\right) - 3 \left(4 {t}^{3}\right) + 0$ $\text{ }$ (we often skip writing this)

$f ' \left(t\right) = 3 {t}^{5} - 12 {t}^{3}$