Question

find the derivatives of trigonometric functions of the following: plss need help for my homework its so difficult

Answer 1

1.\[(2x+3)^{2}\sin(x-6)\]

2.\[y=(\frac{ x+1 }{ x })\tan (\frac{ x+1 }{ x })\]

3. \[y=\tan ^{3}(cosx ^{2})\]

Answer 2

For the first one, we'll need to apply the product rule.\[\large (uv)'=u'v+uv'\]

\[\large \frac{d}{dx}(2x+3)^{2}\sin(x-6) =\]
\[\large \color{royalblue}{\frac{d}{dx}(2x+3)^2} \sin(x-6)+(2x+3)^2 \color{royalblue}{\frac{d}{dx}\sin(x-6)}\]Understand the setup? We'll need to take the derivative of the blue terms.

Answer 3

So looking at the first term,\[\large 2(2x+3)\color{royalblue}{\frac{d}{dx}(2x+3)} \sin(x-6)\]After we apply the power rule, we have to apply the chain rule.

Answer 4

Applying the chain rule gives us this,\[\large 2(2x+3)\color{royalblue}{(2)} \sin(x-6)\]Which simplifies to,\[\large 4(2x+3) \sin(x-6)\]

Answer 5

Then for the second term,\[\large (2x+3)^2 \color{royalblue}{\frac{d}{dx}\sin(x-6)}\]The derivative of sine gives us cosine,\[\large (2x+3)^2 \cos(x-6)\color{royalblue}{\frac{d}{dx}(x-6)}\]Applying the chain rule again shows us that the derivative of the inner function is simply 1.\[\large (2x+3)^2 \cos(x-6)\color{royalblue}{(1)}\]

Answer 6

Giving us an answer of,\[\large 4(2x+3) \sin(x-6)+(2x+3)^2 \cos(x-6)\]

Answer 7

Confused by any of that?
I think setting it up correctly is the hardest part, at least for me.
Things can certainly get messy as you get further through the problem also though.

Answer 8

i understand a little bit anyway tnx