Question

Find the derivative:

Answer 1

\[y=(\theta^2+\sec \theta+1)^3\]

Answer 2

Chain rule comes to mind...

Answer 3

\(\frac{d}{dx} [a]^n = \frac{d}{dx}n[a]^{n-1} \times \frac{d}{dx}(a)\)

Answer 4

or mayb you should think of \(a\) as \(f(x)\) instead of single term \(a\) it kind of looks like power rule

Answer 5

Let's let
\[\Large u = \theta ^2 +\sec\theta + 1\]
So that we have
\[\Large y = u^3\]
Then
\[\Large \frac{dy}{d\theta}=\frac{dy}{du}\cdot \frac{du}{d\theta}\]
as per the chain rule...

Answer 6

\(f(\theta)\)**

Answer 7

Think of this \[\Large y= (\color{blue}{\theta^2 + \sec\theta+1} )^3\]as a single entity first, and differentiate, using the power rule:
\[\Large \frac{d}{dx}x^n = nx^{n-1}\]

Answer 8

so first part:
\[y'=3(\theta^2+\sec \theta+1)^2*2 \theta+secxtanx\]

Answer 9

Better enclose these in parentheses instead...and it's \(\large \sec\color{red}\theta \tan \color{red}\theta\) not x lol...
\[\Large y' = 3(\theta^2 + \sec\theta + 1)^2(2\theta+ \sec\theta\tan\theta )\]
Yes, parentheses matter :3

Answer 10

Just don't drop the parenthesis around (2theta + sectan) and I was beat to it, haha.

Answer 11

Oops, right.. thanks guys :)

Answer 12

\(\huge \theta\)