Question

y = sin (e3x) the dy/dx =

Answer 1

\[\sin(e^{3x})\]?

Answer 2

yes

Answer 3

ok so remember this
(f(g(x))'=f'(g(x)) g'(x)
ok ?

Answer 4

chain rule twice

Answer 5

yep

Answer 6

\[y = \sin(e^{3x})\]
\[\frac{dy}{dx} = \frac{d(\sin(e^{3x}))}{dx}\]
Now use chain rule.
\[\frac{d(\sin(e^{3x})) * d(e^{3x})}{d(e^{3x}) * dx}\]
You'll get. [Use chain rule again in the next step]
\[\cos(e^{3x})\frac{d(e^{3x})*d(3x)}{d(3x)*dx}\]
Then you get:
\[\cos(e^{3x}) ~*~ e^{3x} * 3\]

Getting this? :)

Answer 7

going step by step...

Answer 8

See through the text. I have posted the steps.
Are you ever of this rule called the chain rule?

Answer 9

aware*

Answer 10

No problem. I'll explain again.

Answer 11

ok

Answer 12

the derivative of sine of something is cosine of something times the derivative of something
the derivative of e to the power of something is e to the something times the derivative of something
the derivative of a constant times x is the constant

Answer 13

\[\frac{d}{dx}[\sin(\text {whatever})]=\cos(\text{whatever})\times \text{whatever}'\]

Answer 14

Yes exactly!

\[\frac{d}{dx}[cos(anything)] = ?\]

What do we have here? We have to differentiate \(cos(anything)\) with dx. Is that possible? No. We can only differentiate \(cos(anything)\) with \(d(anything)\) not with dx.

Because \[\frac{d}{d('x')}cos ('x') ~~gives~~us~~ = -sin('x')\]

So why don't we multiply \(d(anything)\) to the numerator AND the denominator to make things right?

So, we'll get:

\[\frac{d}{dx}[cos(anything)] = -sin(anything) * \frac{d}{dx}(anything)\]

[This is the chain rule]

Understood? :)

Answer 15

yes.. thank you.. :-)

Answer 16

By the way,
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