How to find instantaneous rate of change for #y=4x^3+2x-3# at x=2?

2 Answers
Barry H.
May 10, 2018

#50#

Explanation:

We need to differentiate the expression, #4x^3+2x-3# to find the slope of the tangent, #d/dx[4x^3+2x-3]# = #12x^2+2#, [ by the general power rule for differentiation, i.e, if #y=ax^n#, #dy/dx=anx^[n-1]#]
and so when #x=2#, #dy/dx# =#12[2]^2+2# = #50#.

This is the rate of change of #y# with respect to #x# at the point where #x=2#, and means #y# is changing fifty times faster than #x# at this point. Hope this was helpful.

Jacobi J.
May 10, 2018

#50#

Explanation:

"Instantaneous rate of change" is just a fancy way of saying "derivative". We need to differentiate this business and plug in #2# at the end.

We can find the derivative using the power rule. Here, we multiply the constant times the exponent, and the power gets decremented. Doing this, we get

#y'=12x^2+2#

NOTE: Recall that the derivative of an #x# term is just its coefficient, and the derivative of a constant is #0#.

Now, we can plug #2# in for #x# to get

#=12(2)^2+2#

#=48+2#

#=50#

The instantaneous rate of change at #x=2# is #50#.

Hope this helps!

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