
step by step solution -:
The given equation of the curve is:
y=3x^{2}-x^{3}
Let assume that y=f(x), it follows that:
f(x)=3x^{2}-x^{3}
Find the derivative:
\frac{d}{dx}{f(x)}=f'(x)=6x-3x^2
since it is given that the tangent line to the given curve is at point (1,2) so substitute (x,y)=(1,2) into the above equation to find the slope of the tangent, it follows
f'(\textcolor{primary}{1})=6(\textcolor{primary}{1})-3(\textcolor{primary}{1})^{2}\implies f'(1)=3
So the tangent line to the curve has the slope 3 at the point (1,2)
by point-slope form, we know that equation of the line is
(y-y_1)=m(x-x_1)
where,
m = slope of the tangent
(\textcolor{secondary}{x_1},\textcolor{tertiary}{y_1})=(\textcolor{secondary}{1},\textcolor{tertiary}{2}),m=3
on substituting the above expression we get
y-\textcolor{tertiary}{2}=3\cdot(x-\textcolor{secondary}{1})
after simplifying the above expression we get
3x-y=1
hence, the required equation of the tangent is: 3x-y=1
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