Evaluate the derivative of the function below at the point (, ). y=7/x + sqrt cos(x)
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well the derivative of that function is just -7/(x^2)+[1/(2(sqrt(cos(x)))]*-sin(x) evaluation of derivative of sqrt[cos(x)] is just applying the chain rule. now all u have to do is substitute the value of x in.
This is the part I don't get...actually putting TT/4 into the equation... I know cos(x) = sqrt2/2 and the same for sin (x), right?
Originally Posted by tradar Evaluate the derivative of the function below at the point (, ). y=7/x + sqrt cos(x) First, rewrite the function like this: then, by the chain rule: and : is not a critical numberthe curve is falling at
Last edited by sinewave85; Mar 21st 2009 at 06:54 AM.
Originally Posted by tradar Evaluate the derivative of the function below at the point (, ). y=7/x + sqrt cos(x)
Originally Posted by tradar This is the part I don't get...actually putting TT/4 into the equation... I know cos(x) = sqrt2/2 and the same for sin (x), right? Right. Just to clarify:
Last edited by sinewave85; Mar 21st 2009 at 06:55 AM.
Originally Posted by sinewave85 Right. Just to clarify: I think there is a mistake!
Originally Posted by u2_wa I think there is a mistake! Big oops. Thanks. I somehow lost that x as I went along.
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