How do you find the slope of the tangent line to the parabola $y=7x-{{x}^{2}}$ at the point (1,6)?

Hint: In this question, we are given an equation of parabola and we need to find the slope of tangent to this curve at point (1,6). For this, we need to calculate the derivative of a given equation with respect to x. We will get the function $\dfrac{dy}{dx}$ which is the slope of tangent at any point on the curve. For finding the slope of tangent at specific point $\left( {{x}_{1}},{{y}_{1}} \right)$ we need to put the value of x only as ${{x}_{1}}$ in the $\dfrac{dy}{dx}$ and find the slope. We will get ${{\left. \dfrac{dy}{dx} \right]}_{x={{x}_{1}}}}=m$ where m is the slope. Derivative of ${{x}^{n}}$ is given by $n{{x}^{n-1}}$.

Complete step by step answer:
Here we are given the equation of parabola as $y=7x-{{x}^{2}}$. We need to find the slope of the tangent at point (1,6). For this, let us first calculate the slope of the tangent at any point for the given curve. We need to find $\dfrac{dy}{dx}$.
The curve is given as $y=7x-{{x}^{2}}$.
Taking derivative with respect to x on both sides we get $\dfrac{dy}{dx}=\dfrac{d}{dx}\left( 7x-{{x}^{2}} \right)$.
We know that $\dfrac{d}{dx}{{x}^{n}}$ is equal to $n{{x}^{n-1}}$. So our equation becomes $\dfrac{dy}{dx}=7{{x}^{1-1}}-2{{x}^{2-1}}=7{{x}^{0}}-2{{x}^{1}}\Rightarrow \dfrac{dy}{dx}=7-2x\left( \text{because }{{x}^{0}}=1 \right)$
As we know, the slope of tangent is given by $y=7x-{{x}^{2}}$. So the slope of tangent at any point on the curve $\dfrac{dy}{dx}$ will be given by 7-2x.
We need to find the slope of the tangent at (1,6). Here, x coordinate is 1. Therefore, the slope of the tangent at x = 1 will be given by putting x = 1 in the value of $\dfrac{dy}{dx}$. We get, ${{\left. \dfrac{dy}{dx} \right]}_{x={{x}_{1}}}}=7-2\left( 1 \right)=7-2=5$.
Hence the slope of the tangent at (1,6) on the parabola $y=7x-{{x}^{2}}$ is equal to 5.
Therefore, the required slope is equal to 5.

Note: Students should note that, when finding slope at any point, we just need the value of x coordinate to get the required slope. Keep in mind the formula of finding derivatives. In general, the derivative of $y=a{{x}^{2}}+bx+c$ is $y'=2ax+b$. Here $y'=\dfrac{dy}{dx}$.