Question

How do you find the slope of the tangent line to a curve at a point?

Answer 1

Hint: To solve this problem we will apply the definition of the derivative; \[\displaystyle \lim_{\Delta h \to 0}\left( \dfrac{f\left( x+\Delta h \right)-f\left( x \right)}{\left( x+\Delta h \right)-\left( x \right)} \right)\]it is clear that derivative of a function is the slope of the function. Therefore, the slope of the tangent line to a curve at a point is the derivative of a function at that point. So, to solve this question, we will, first of all, determine the value of y corresponding to the value of x by substituting x into the equation, then we differentiate the function at the point $ \left( x,y \right) $ to find the slope to the tangent line to the curve.

Complete step by step answer:
From the problem, we are asked how we could find the slope of the tangent line to a curve at a point.
The tangent line to a curve is a straight line that touches the curve at that point.
Now, let us assume a function $ f\left( x \right) $ for the curve, to find the slope of the tangent line to a curve at a point, first we find the value of $ f\left( x \right) $ corresponding to x by substituting the value of x into $ f\left( x \right) $ . Then we differentiate \[f\left( x \right)\] at this point using the definition of the derivative\[\displaystyle \lim_{\Delta h \to 0}\left( \dfrac{f\left( x+\Delta h \right)-f\left( x \right)}{\left( x+\Delta h \right)-\left( x \right)} \right)\].
The value of the derivative of \[f\left( x \right)\] at the point $ \left( x,f\left( x \right) \right) $ is the slope of the tangent line to the curve at that point.
For example if we intend to find the slope of the tangent line to a curve \[f\left( x \right)=3{{x}^{2}}+2x-1\] at $ x=1 $ .
First, we substitute the value of $ x=1 $ into \[f\left( x \right)=3{{x}^{2}}+2x-1\] , we get:
\[f\left( 1 \right)=3+2-1=4\]
Then, we differentiate \[f\left( x \right)=3{{x}^{2}}+2x-1\] at the point $ \left( 1,4 \right) $ , differentiating \[f\left( x \right)=3{{x}^{2}}+2x-1\] with respect to x, we get; \[f'\left( x \right)=6x+2\]
Now, we can go ahead and substitute the value of x=1 into \[f'\left( x \right)=6x+2\]
\[\therefore f'\left( 1 \right)=6+2=8\]
Hence, the slope of the tangent line to the curve \[f\left( x \right)=3{{x}^{2}}+2x-1\] at $ x=1 $ is 8.

Note:
 It is important to note that care should be taken when making the necessary substitutions and when finding the derivative of the function. For a tangent line between two points $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ , the slope of the tangent line is given as \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]. Also, if we need to derive an equation of the tangent line to a curve at a point, we simply substitute the values of \[x,f\left( x \right)\] and $ f'\left( x \right) $ into $ f\left( x \right)=f'\left( x \right)x+c $ to find the $ f(x) $ intercept c and then we go ahead to substitute the value of c and the value of the slope of the tangent m into the equation of a straight line $ y=mx+c $ . The resulting equation is that of the tangent line to \[f\left( x \right)\] at that point. In a hurry sometimes we might end up finding the derivative of \[f\left( a \right)\] instead of \[f\left( x \right)\] which gives us 0 as the only derivative in each case. So, we have to take care of this point as well.