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OpenStudy (anonymous):
Find the derivative of f(x) = -cos (2x^2-3)
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OpenStudy (anonymous):
Do you know how to use the chain rule?
OpenStudy (anonymous):
yeah yes i do
OpenStudy (anonymous):
Well your inner function is \(2x^2-3\) and your outer function is \(-\cos(x)\). There shouldn't be anything hard about this.
OpenStudy (anonymous):
do we have to use chain rule ? i'm confused on what we should do when with the -cos
OpenStudy (anonymous):
The - is just \(-1\cdot \cos(x)\). It's a constant. You can pull it out and put it back in after.
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OpenStudy (anonymous):
oh really? but this have to use f(x+h)-f(x) first derivation ..
OpenStudy (anonymous):
You have to use this you mean? \[\Large \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
OpenStudy (anonymous):
You're not allowed to use the chain rule?
OpenStudy (anonymous):
Oh sorry, i just realised it only has to use chain rule. I make things go more complicated -_-
OpenStudy (anonymous):
So can you do it now?
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OpenStudy (anonymous):
yeah the answer is sin(2x^2-3)(4x) right ?
OpenStudy (raden):
yeah, i agree that
OpenStudy (anonymous):
Oh thank you so much :D
OpenStudy (anonymous):
It's correct.
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