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First of all there is an equation

Then there is the derivative

Then there is a point slope formula to find the equation of the tangent line

Point slope formula to obtain the tangent line .

y=3a^{2}(x-a)+a^{3}

Then Plug in the x coordinate into the derivative to get the slope

f'(1) = 3(1)^{2}

f'(1) = 3

What this means is that for any value of x=a, the instantaneous slope of f at (a,a^{3}) is 3a^{2}.

Here , i don't really understand some change of terms from f(x) to y , a ... etc

Please help

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# B Some doubts about these differentiation steps

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