Question #ccf19

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Answer 1 · 2017-07-17T21:01:12

We will use the chain rule.

so we must solve for #t# and we get #x=t^2 iff t=sqrtx#
so #dy/dx=dy/dtdt/dx=2*1/(2sqrtx)=1/sqrtx#

Answer 2 · 2017-07-18T09:44:37

#dy/dx=1/t#

#•color(white)(x)dy/dx=(dy/dt)/(dx/dt)#

#y=2t+3rArrdy/dt=2#

#x=t^2rArrdx/dt=2t#

#rArrdy/dx=2/(2t)=1/t=t^-1#