The speed(v) of a particle moving along a straight line is given by `v=(t^(2)+3t-4` where v is in m/s and t in seconds. Find time t at which the particle will momentarily come to rest.
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(i) Here `f(x)=sinx, g(x)=x` So, `f'(x)=cos x, g'(x)=1` Therefore `(dy)/(dx)=((cos x)(x)-(sin x)(1))/x^(2)=(x cos x-sin x)/x^(2)` (ii) Here `f(x)=4x^(3), g(x)=e^(x)` So `f'(x)=12x^(2), g'(x)=e^(x)` Therefore, `(dy)/(dx)=(12x^(2)(e^(x))-4x^(3)(e^(x)))/((e^(x))^(2))=(12x^(2)-4x^(3))/e^(x)`
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