How could I find the equation of the tangent to the graph for this function: `f(x)= sin(e^(3*x^(2) -12)-1) + 2x` to the point (2,4) ?



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Posted on (Answer #1)

Equation of the tangent to the graph of function `f` at point `(x_0,f(x_(0) ) )` is

`y-f(x_0)=f'(x_0)(x-x_0)`                                   (1)

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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