Derivative of y=4sec5x

[Sophie]

Hello if anyone could show the work to an exercise question I have. I have the question and the answer breifly broken down, however I am not sure where the 4 comes into play.

Question.

What is the Derivative of y=4sec5x

Answer 20sec5xtan5x

I thought there would be three parts to this,
y=f(u)
u=g(x)
x=h(t)
but from the anser there appears to be only 2, could someone explain what the inner and outer function is and why there are only 2.

I would be very greatful for any help, Thanks Sophie

 

[soroban]

Hello, Sophie!

You're making awfully hard work out of it . . .

What is the derivative of: \(\displaystyle \, y\:=\:4\cdot\sec(5x)\)

The "4" is just a coefficient . . . you just "bring it along". **

The outer function is the \(\displaystyle sec\), the inner function is \(\displaystyle 5x\).

So we have: \(\displaystyle \;y'\:=\:4\,\cdot\,\sec(5x)\tan(5x)\,\cdot\,5 \:=\:20\cdot\sec(5x)\tan(5x)\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

If you want to make hard work for yourself, consider: \(\displaystyle y\:=\:5x^3\)

You may ask, "What do I do about the 5?"

Well, you have a product: \(\displaystyle \:y\:=\5)\cdot(x^3}\)

. . We can use the Product Rule: \(\displaystyle \:y'\:=\5)(3x^2)\,+\,(0)(x^3) \:=\:15x^2\)

But do we do that every time? . . . Certainly not!

We "bring along" the 5 . . . \(\displaystyle y' \:=\:5(3x^2) \:=\:15x^2\) . . . right?

 

[Sophie]

Thank you very much

I knew it was simple, I need to understand the loger version 1st to be able to make simple work of it.

Sophie