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Equation of the tangent to the graph of function `f` at point `(x_0,f(x_(0) ) )` is
So we first need to find derivation of `f`. For this we will need chain rule `(f(g(x)))'=f'(g(x))g'(x)`.
`f'(x)=cos(e^(3x^2-12)-1)e^(3x^2-12) cdot 6x+2`
Since we already know that `f(2)=4` we can put all this into formula (1).
So equation of tangent line of graph of function `f` at point `(2,4)` is `y=14x-24`.
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