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d/dx csc(x)

= - csc(x) tan(x)

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โˆ™ 2011-09-07 02:50:04
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Q: What is the derivative of cscx?
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How do you take the derivative of a trig function?

Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x


How do you find the derivative of - csc x - sin x?

d/dx (-cscx-sinx)=cscxcotx-cosx


What is the anti-derivative of co secant x?

According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.


What is the derivative of -cscxcotx?

Using the u substitution method of derivation (selecting sinx as u and cosxdx as du), you get f'(x)=cscx.


What is the integral of cscx?

- ln (cscx + cotx) + C You use u substitution.


Can you use the equation given below to find the second derivative of pi divided by 6 if fx equals cscx?

pi divided by 6 is a constant and so its first derivative is 0. And since that is also a constant, the second derivative is 0. It is not clear what f(x) = csc(x) has to do with that!


What is the anti derivative of the square root of 1-x2?

-1


How do you prove that the derivative of csc x is equals to -csc x cot x?

d/dx cscx = d/dx 1/sinx = d/dx (sinx)-1= -(sinx)-2 cosx = -cosx/sin2x = -1/sinx.cosx/sinx = -cscx cotx I suggest you copy this out onto paper so it is more clear. The / signs make it harder to see what is happening compared to horizontal divide lines.


Is it derivative of or derivative from?

"Derivative of"


How do you verify the identity sinx cscx 1?

sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.


What is the derivative of 1 divided by sinx?

y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx


What is the second derivative of a function's indefinite integral?

well, the second derivative is the derivative of the first derivative. so, the 2nd derivative of a function's indefinite integral is the derivative of the derivative of the function's indefinite integral. the derivative of a function's indefinite integral is the function, so the 2nd derivative of a function's indefinite integral is the derivative of the function.

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