fry10fry10
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- Sep 22, 2009
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I need help calculating the derivative of 6e^x/sinx. I believe it is [sinx * 6e^x - sinx * cosx]/sin^2(x). Is this correct, can I simplify this further?
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calculating derivative of 6e^x/sinx
- Thread starter fry10fry10
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One way to go would be to rewrite as \(\displaystyle 6e^{x}csc(x)\) and use the product rule instead of the quotient rule.
\(\displaystyle \frac{d}{dx}[e^{x}csc(x)]=e^{x}(-csc(x)cot(x))+e^{x}csc(x)=\frac{e^{x}(sin(x)-cos(x))}{sin^{2}(x)}\)
Just another way. Just don't forget the 6.
\(\displaystyle \frac{d}{dx}[e^{x}csc(x)]=e^{x}(-csc(x)cot(x))+e^{x}csc(x)=\frac{e^{x}(sin(x)-cos(x))}{sin^{2}(x)}\)
Just another way. Just don't forget the 6.
fry10fry10
New member
- Joined
- Sep 22, 2009
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galactus said:One way to go would be to rewrite as \(\displaystyle 6e^{x}csc(x)\) and use the product rule instead of the quotient rule.
\(\displaystyle \frac{d}{dx}[e^{x}csc(x)]=e^{x}(-csc(x)cot(x))+e^{x}csc(x)=\frac{e^{x}(sin(x)-cos(x))}{sin^{2}(x)}\)
Just another way. Just don't forget the 6.
Thanks I appreciate it.