If the normal to the curve y=f(x) at the poin | vedclass

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(D) The slope of the normal to the curve $y=f(x)$ at a point is given by $	an 	heta$,where $	heta$ is the angle the normal makes with the positive

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$1$

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(D) The slope of the normal to the curve $y=f(x)$ at a point is given by $	an 	heta$,where $	heta$ is the angle the normal makes with the positive $X$-axis.Given $	heta = \frac{3 \pi}{4}$,the slope of the normal is $	an \left(\frac{3 \pi}{4}ight) = -1$.We know that the slope of the normal is related to the slope of the tangent $\left(\frac{dy}{dx}ight)$ by the formula: $	ext{Slope of normal} = -\frac{1}{\frac{dy}{dx}}$.Substituting the values,we get $-1 = -\frac{1}{f^{\prime}(3)}$.Solving for $f^{\prime}(3)$,we get $f^{\prime}(3) = 1$.

If the normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle of $\frac{3 \pi}{4}$ with the positive $X$-axis,then the value of $f^{\prime}(3)$ is

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