Question

Find the slope of the tangent line to the graph of the function at the given point.
g(x)=5-x^2, (2,1)

Answer 1

Topic :-

Differentiability

Given :-

g(x) = 5 - x²

To Find :-

The slope of the tangent line to the graph of the given function at point (2, 1).

Methodology :-

In such type of questions, we find derivative of given function as derivative of a function gives us slope. After that we put the value of coordinates of point to which tangent is drawn in the obtained derivative of the function.

Solution :-

Given Function :-

g(x) = 5 - x²

Derivative of the function :-

$\sf {\dfrac{d(g(x))}{dx}=\dfrac{d(5-x^2)}{dx}}$

$\sf {\dfrac{d(g(x))}{dx}=\dfrac{d(5)}{dx}-\dfrac{d(x^2)}{dx}}$

$\sf {\left( \because \dfrac{d(f\pm g)}{dx}=\dfrac{d(f)}{dx}\pm\dfrac{d(g)}{dx}\right)}$

$\sf {\dfrac{d(g(x))}{dx}=0-\dfrac{d(x^2)}{dx}}$

$\sf {\left( \because \dfrac{d(k)}{dx}=0,where\:k\:is\:constant.\right)}$

$\sf {\dfrac{d(g(x))}{dx}=-2x^{2-1}}$

$\sf {\left( \because \dfrac{d(x^n)}{dx}=nx^{n-1}\right)}$

$\sf {\dfrac{d(g(x))}{dx}=-2x}$

So, we have obtained the derivative of the function.

Calculating Slope at the given point,

Put x = 2 and y = 1 in the obtained derivative.

$\sf {\dfrac{d(g(x))}{dx}=-2(2)}$

$\sf {\dfrac{d(g(x))}{dx}=-4}$

Note : There were no terms of 'y' in the obtained derivate. So, we have not put y = 1 anywhere in obtained derivative of the given function.

Answer :-

The slope of the tangent line to the graph of the given function at point (2, 1) is -4.