Slope of the curve = slope of the tangent line draw at the particular point
To find slope of the curve at a given point, we have to follow the steps given below.
Step 1 :
Find the derivative of the given function using the appropriate rule.
Step 2 :
Apply the given point (x, y) in the first derivative function.
Step 3 :
Do the possible simplification, after the simplification done the result is the slope of the tangent line drawn at the particular point.
Find the gradient of the tangent to
Problem 1 :
f(x) = 3x2 at x = -1
Solution :
f(x) = 3x2
f'(x) = 3(2x)
= 6x
Apply x = -1 in the first derivative.
f'(-1) = 6(-1)
= -6
So, -6 is the slope of the tangent line for the curve drawn at the point x = -1
Problem 2 :
f(x) = 6/x at x = 2
Solution :
f(x) = 6/x
Here x is at the denominator, we can move the x to the numerator and find the derivative.
f(x) = 6x-1
f'(x) = -6x-2
f'(x) = -6/x2
Slope at x = 2
f'(2) = -6/22
= -6/4
= -3/2
So, the required slope is -3/2.
Problem 3 :
f(x) = x2 + 2x + 1 at x = 1
Solution :
f(x) = x2 + 2x + 1
f'(x) = 2x + 2(1) + 0
= 2x + 2
Slope at x = 1
f'(1) = 2(1) + 2
= 2 + 2
f'(1) = 4
So, the required slope is 4.
Problem 4 :
f(x) = x2 + 7x at x = -2
Solution :
f(x) = x2 + 7x
f'(x) = 2x + 7(1)
= 2x + 7
Slope at x = -2
f'(-2) = 2(-2) + 7
= -4 + 7
f'(-2) = 3
So, the required slope is 3.
Problem 5 :
f(x) = (x2 + 1)/x at x = 2
Solution :
f(x) = (x2 + 1)/x at x = 2
Here the variable x is at both numerator and in the denominator, so we have to use quotient rule to find the derivative.
u = x2 + 1 and v = x
u' = 2x + 0 ==> 2x, v' = 1
Applying the quotient rule,
d(u/v) = (vu' - uv')/v2
= [x(2x) - (x2 + 1)(1)]/x2
= (2x2 - x2 - 1)/x2
f'(x) = (x2 - 1)/x2
Slope at x = 2
f'(2) = (22 - 1)/22
= (4 - 1) / 4
= 3/4
So, the required slope is 3/4.
Problem 6 :
f(x) = √x + (8/x) at x = 4
Solution :
f(x) = √x + (8/x)
f(x) = √x + (8x-1)
f'(x) = (1/2√x) - (8/x2)
Slope at x = 4
f'(4) = (1/2√4) - (8/42)
= 1/2(2) - (8/16)
= 1/4 - 1/2
= (1 - 2)/4
= -1/4
So, the required slope is -1/4.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM