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Mathematics 15 Online
OpenStudy (anonymous):

Use the Quotient Rule to differentiate the function. MEDAL WILL BE REWARDED! g(t)=(t^2+2)/(2t-7)

OpenStudy (anonymous):

(2t)(2t-7)-(t^2+2)(2)/(2t-7)^2

OpenStudy (anonymous):

\[g(t)=\frac{(t^2+2)}{(2t-7)}\] \[ g'(t)=\frac{ (2t-7) \frac{d}{dx}(t^2+2)- (t^2+2) \frac{d}{dx}(2t-7)}{(2t-7)}\] \[\rightarrow g'(t)=\frac{ (2t-7)(2t+0)- (t^2+2) (2 \times 1-0)}{(2t-7)^2}\] \[\rightarrow g'(t)=\frac{ (2t-7) 2t- (t^2+2) 2}{(2t-7)^2}\] \[\rightarrow g'(t)=\frac{ 4t^2-14t- 2t^2-4}{(2t-7)^2}\] \[\rightarrow g'(t)=\frac{ 2t^2-14t-4}{(2t-7)^2}\] is the required differentiation. @CarlyLiberty

OpenStudy (anonymous):

Thank you that's exactly what I got! @dpasingh

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