`f(x)=2^(3x^4sin(3x))`

Use the chain rule and `(d)/(dx) a^x = ln(a)a^x`

`f'(x)=ln(2)2^(3x^4sin(3x))*(d)/(dx)(3x^4 sin(3x))`

Using the product property and `(d)/(dx)3x^4 = 12x^3` and `(d)/(dx)sin(3x)=3cos(3x)` we get

`f'(x)=ln(2)2^(3x^4sin(3x))(12x^3sin(3x)+3x^4(3cos(3x)))`

Simplifying we get

`f'(x)=3ln(2)x^3(4sin(3x)+3cos(3x))2^(3x^4sin(3x))`

`m=f'(1)=3ln(2)(4sin(3)+3cos(3))2^(3sin(3)) `

f(1)=m(1)+b in order to be tangent to the curve we get

`b = f(1)-m = 2^(3sin(3))-3ln(2)(4sin(3)+3cos(3))2^(3sin(3))`

Simplifying we get

`b = (1-3ln(2)(4sin(3)+3sin(3)))2^(3sin(3))` and

` m=3ln(2)(4sin(3)+3cos(3))2^(3sin(3))`

The equation we get is

`y=3ln(2)(4sin(3)+3cos(3))2^(3sin(3))x+(1-3ln(2)(4sin(3)+3cos(3))2^(3sin(3))`

Or approximately

`y=-6.70804063 x + 8.04908783`

Looks good.

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