You follow the steps given below to find the slope of a tangent line to a curve at a given point using derivative.
Step 1 :
Let y = f(x) be the function which represents a curve. Find the derivative ᵈʸ⁄dₓ or f'(x), where ᵈʸ⁄dₓ or f'(x) is the slope of the line tangent to the curve at any point.
Step 2 :
Substiute the given point into ᵈʸ⁄dₓ or f'(x) and evaluate.
In each case, find the slope of a line tagent to the graph of the function at the given point.
Example 1 :
f(x) = 3  5x, (1, 8)
Solution :
f(x) = 3  5x
f'(x) = 5(1)
f'(x) 5
Slope of the tangent line at (1, 8) :
m = 5
Example 2 :
Solution :
Slope of the tangent line at (2, 2) :
Example 3 :
f(x) = 2x^{2}  3, (2, 5)
Solution :
f(x) = 2x^{2}  3
f'(x) = 2(2x)  0
f'(x) = 4x
Slope of the tangent line at (2, 5) :
m = 4(2)
m = 8
Example 4 :
f(x) = 5  x^{2}, (3, 4)
Solution :
f(x) = 5  x^{2}
f'(x) = 0  2x
f'(x) = 2x
Slope of the tangent line at (3, 4) :
m = 2(3)
m = 6
Example 5 :
f(t) = 3t  t^{2}, (0, 0)
Solution :
f(t) = 3t  t^{2}
f'(t) = 3(1)  2t
f'(t) = 3  2t
Slope of the tangent line at (0, 0) :
m = 3  2(0)
m = 3
Example 6 :
h(t) = t^{2} + 4t, (1, 5)
Solution :
h(t) = t^{2} + 4t
h'(t) = 2t  4(1)
h'(t) = 2t  4
Slope of the tangent line at (1, 5) :
m = 2(5)  4
m = 10  4
m = 6
Example 7 :
x^{2} + x^{2} = 25, (6, 3)
Solution :
x^{2} + x^{2} = 25
Find the derivative on both sides with respect to x.
Slope of the tangent line at (6, 3) :
Example 8 :
x = 3t and y = t^{2} + 1, t = 2.
Solution :


Slope of the tangent line at t = 2 :
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