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Mathematics 7 Online
OpenStudy (anonymous):

If f(x)=x4+2cos(x) / 2sin(x), find f′(π/2).

OpenStudy (amistre64):

whats our derivative?

OpenStudy (anonymous):

f(x) = 4x^3 + (-2sinx) (2sinx) + (x^4 +cosx) ( 2cosx) / 2sinx^2

OpenStudy (amistre64):

f(x)=x^4+2cos(x) / 2sin(x) f(x)=x^4 csc(x) + cot(x) f(x)= 4x^3 csc(x) -x^4 cot(x)csc(x) - csc^2(x)

OpenStudy (amistre64):

i forgot the /2 on the first term doh

OpenStudy (anonymous):

\[f \left( x \right)=\frac{ x^4 +2\cos \left( x \right) }{ 2\sin \left( x \right) }\,\, \text{ or } f \left( x \right)= x^4 +\frac{2\cos \left( x \right) }{ 2\sin \left( x \right) }\]

OpenStudy (amistre64):

f(x)= 2x^3 csc(x) -x^4 cot(x)csc(x)/2 - csc^2(x)

OpenStudy (amistre64):

csc(pi/2) = sin(pi/2) = 1 sooo

OpenStudy (amistre64):

isnt cot(pi/2) = 0 since its perp to a vertical slope?

OpenStudy (anonymous):

i don't know much about trig

OpenStudy (amistre64):

then youll need more practice. trig id 98% memorization

OpenStudy (amistre64):

** trig is ....

OpenStudy (anonymous):

why did you use csc?

OpenStudy (amistre64):

tangent is the slope of a line thru the center of a circle .... pi/2 is 90 degrees is attributed to 1/0 cotangent is the reciprocal of tangent giving us 0/1

OpenStudy (amistre64):

1/sin=csc

OpenStudy (amistre64):

instead of the quotient rule, i just broke it all apart and used trig identities

OpenStudy (anonymous):

okay thx i think i got it now!

OpenStudy (amistre64):

yw, my work of it my have some errors so i hope you dont use it verbatim

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