What is called slope of tangent?

It is the derivative of a function at a point, which represents the steepness of the curve at that exact point.

Can you find the slope of a tangent line?

Yes! Take the derivative of the function and substitute the given xx-value.

What is the slope of this tangent?

It depends on the curve. Find the derivative of the curve and evaluate it at the given point.

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Slope of Tangent

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Slope of Tangent

In calculus and geometry, a tangent is a straight line that just “touches” a curve at a single point without crossing it. This line represents the instantaneous rate of change of the function at that point. In simple terms, the slope of the tangent tells us how steep the curve is at that point.

1.0What is the Slope of a Tangent?

The slope of a tangent line to a curve at a given point is the derivative of the function at that point.

In other words:

Slope of tangent = Derivative of the function at the point

This slope gives the direction of the tangent line and helps determine whether the function is increasing or decreasing at that point.

2.0What is the Slope of Tangent Formula?

For a function y = f(x), the slope of the tangent line at x = a is given by:

Slope of Tangent = f′(a)

Where:

f'(x) is the first derivative of the function
a is the x-coordinate of the point of tangency

3.0Tangent Line Equation

Once the slope of the tangent is known, we can find the equation of the tangent line using the point-slope form:

$y - y_{1} = m (x - x_{1})$

Where:

$(x_{1}, y_{1})$ is the point on the curve
$m = f^{'} (x_{1})$ is the slope of the tangent

4.0How to Find the Slope of a Tangent Line?

Step by step:

Find the derivative f'(x) of the function y = f(x).
Evaluate f'(x) at the required point x = a.
The result is the slope of the tangent at that point.

5.0Solved Examples on Slope of Tangent

Example 1: Find the slope of the tangent to the curve

$y = x^{3} - 5 x + 4 at x = 2$

Solution:

$We find the derivative: \frac{d y}{d x} = 3 x^{2} - 5 A t x = 2 : \frac{d y}{d x} = 3 (2)^{2} - 5 = 12 - 5 = 7 Slope = 7$

Example 2: Find the slope of the tangent to the curve

$y = e^{x} + 3 x at x = 0$

Solution:

$\frac{d y}{d x} = \frac{d}{d x} (e^{x} + 3 x) = e^{x} + 3 A t x = 0 : \frac{d y}{d x} = e^{0} + 3 = 1 + 3 = 4 Sl o p e = 4$

Example 3: Find the slope of the tangent to $y = sin (x) + cos (x) at x = \frac{π}{4}$

Solution:

$\frac{d y}{d x} = cos (x) - sin (x) A t x = \frac{π}{4} : cos (\frac{π}{4}) = \frac{2}{2}, sin (\frac{π}{4}) = \frac{2}{2} \frac{d y}{d x} = \frac{2}{2} - \frac{2}{2} = 0 Slope = 0 (Horizontal Tangent)$

Example 4: Find the slope of the tangent to $y = \frac{x ^{2} + 1}{x} at x = 1$

Solution:

$\frac{d y}{d x} = \frac{x \cdot 2 x - ( x ^{2} + 1 ) \cdot 1}{x ^{2}} \frac{d y}{d x} = \frac{2 x ^{2} - x ^{2} - 1}{x ^{2}} \frac{d y}{d x} = \frac{x ^{2} - 1}{x ^{2}} A t x = 1 : \frac{d y}{d x} = \frac{1 - 1}{1} = 0 Slope = 0$

Example 5: A curve is given parametrically as $x = a cos^{3} θ, y = a sin^{3} θ$ . Find the slope of the tangent at $θ = \frac{π}{4}$

Solution:

$From parametric equations, slope is: \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} We differentiate: x = a cos^{3} θ \Rightarrow \frac{d x}{d θ} = a \cdot 3 cos^{2} θ \cdot (- sin θ) \Rightarrow \frac{d x}{d θ} = - 3 a cos^{2} θ sin θ y = a sin^{3} θ \Rightarrow \frac{d y}{d θ} = a \cdot 3 sin^{2} θ \cdot cos θ \Rightarrow \frac{d y}{d θ} = 3 a sin^{2} θ cos θ \frac{d y}{d x} = \frac{3 a s i n ^{2} θ c o s θ}{- 3 a c o s ^{2} θ s i n θ} - \frac{s i n θ}{c o s θ} = - tan θ A tθ = \frac{π}{4} : \frac{d y}{d x} = - tan (\frac{π}{4}) = - 1 Slope = -1$

Example 6: For the curve defined by $x^{2} + x y + y^{2} = 7$ . Find the slope of the tangent at the point (1,2).

Solution:

$Differentiate both sides implicitly: \frac{d}{d x} (x^{2} + x y + y^{2}) = \frac{d}{d x} (7) Apply derivatives: 2 x + (x \cdot \frac{d y}{d x} + y) + 2 y \cdot \frac{d y}{d x} = 0 \Rightarrow 2 x + y + x \cdot \frac{d y}{d x} + 2 y \cdot \frac{d y}{d x} = 0 G ro u p \frac{d y}{d x} : x \cdot \frac{d y}{d x} + 2 y \cdot \frac{d y}{d x} = - (2 x + y) \Rightarrow \frac{d y}{d x} = \frac{- ( 2 x + y )}{x + 2 y} A t (1, 2) : \frac{d y}{d x} = \frac{- ( 2 \cdot 1 + 2 )}{1 + 2 \cdot 2} = \frac{- 4}{5} Slope = \frac{- 4}{5}$

6.0Practice Problems on Slope of Tangent

Find the slope of the tangent to $y = ln (x^{2} + 1)$ at x = 1.
Find the slope of the tangent to the curve $x^{2} + y^{2} = 25$ at the point (3, 4).
Find the equation of the tangent line to $y = tan x$ at $x = \frac{π}{6} .$
Parametric: $x = t^{2}, y = t^{3} .$ Find slope at t = 2.
Find the slope of the tangent to the curve $x^{3} + y^{3} = 3 a x y$ at point (a, a).

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Slope of Tangent

In calculus and geometry, a tangent is a straight line that just “touches” a curve at a single point without crossing it. This line represents the instantaneous rate of change of the function at that point. In simple terms, the slope of the tangent tells us how steep the curve is at that point.

1.0What is the Slope of a Tangent?

The slope of a tangent line to a curve at a given point is the derivative of the function at that point.

In other words:

Slope of tangent = Derivative of the function at the point

This slope gives the direction of the tangent line and helps determine whether the function is increasing or decreasing at that point.

2.0What is the Slope of Tangent Formula?

For a function y = f(x), the slope of the tangent line at x = a is given by:

Slope of Tangent = f′(a)

Where:

f'(x) is the first derivative of the function
a is the x-coordinate of the point of tangency

3.0Tangent Line Equation

Once the slope of the tangent is known, we can find the equation of the tangent line using the point-slope form:

$y - y_{1} = m (x - x_{1})$

Where:

$(x_{1}, y_{1})$ is the point on the curve
$m = f^{'} (x_{1})$ is the slope of the tangent

4.0How to Find the Slope of a Tangent Line?

Step by step:

Find the derivative f'(x) of the function y = f(x).
Evaluate f'(x) at the required point x = a.
The result is the slope of the tangent at that point.

5.0Solved Examples on Slope of Tangent

Example 1: Find the slope of the tangent to the curve

$y = x^{3} - 5 x + 4 at x = 2$

Solution:

$We find the derivative: \frac{d y}{d x} = 3 x^{2} - 5 A t x = 2 : \frac{d y}{d x} = 3 (2)^{2} - 5 = 12 - 5 = 7 Slope = 7$

Example 2: Find the slope of the tangent to the curve

$y = e^{x} + 3 x at x = 0$

Solution:

$\frac{d y}{d x} = \frac{d}{d x} (e^{x} + 3 x) = e^{x} + 3 A t x = 0 : \frac{d y}{d x} = e^{0} + 3 = 1 + 3 = 4 Sl o p e = 4$

Example 3: Find the slope of the tangent to $y = sin (x) + cos (x) at x = \frac{π}{4}$

Solution:

$\frac{d y}{d x} = cos (x) - sin (x) A t x = \frac{π}{4} : cos (\frac{π}{4}) = \frac{2}{2}, sin (\frac{π}{4}) = \frac{2}{2} \frac{d y}{d x} = \frac{2}{2} - \frac{2}{2} = 0 Slope = 0 (Horizontal Tangent)$

Example 4: Find the slope of the tangent to $y = \frac{x ^{2} + 1}{x} at x = 1$

Solution:

$\frac{d y}{d x} = \frac{x \cdot 2 x - ( x ^{2} + 1 ) \cdot 1}{x ^{2}} \frac{d y}{d x} = \frac{2 x ^{2} - x ^{2} - 1}{x ^{2}} \frac{d y}{d x} = \frac{x ^{2} - 1}{x ^{2}} A t x = 1 : \frac{d y}{d x} = \frac{1 - 1}{1} = 0 Slope = 0$

Example 5: A curve is given parametrically as $x = a cos^{3} θ, y = a sin^{3} θ$ . Find the slope of the tangent at $θ = \frac{π}{4}$

Solution:

$From parametric equations, slope is: \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} We differentiate: x = a cos^{3} θ \Rightarrow \frac{d x}{d θ} = a \cdot 3 cos^{2} θ \cdot (- sin θ) \Rightarrow \frac{d x}{d θ} = - 3 a cos^{2} θ sin θ y = a sin^{3} θ \Rightarrow \frac{d y}{d θ} = a \cdot 3 sin^{2} θ \cdot cos θ \Rightarrow \frac{d y}{d θ} = 3 a sin^{2} θ cos θ \frac{d y}{d x} = \frac{3 a s i n ^{2} θ c o s θ}{- 3 a c o s ^{2} θ s i n θ} - \frac{s i n θ}{c o s θ} = - tan θ A tθ = \frac{π}{4} : \frac{d y}{d x} = - tan (\frac{π}{4}) = - 1 Slope = -1$

Example 6: For the curve defined by $x^{2} + x y + y^{2} = 7$ . Find the slope of the tangent at the point (1,2).

Solution:

$Differentiate both sides implicitly: \frac{d}{d x} (x^{2} + x y + y^{2}) = \frac{d}{d x} (7) Apply derivatives: 2 x + (x \cdot \frac{d y}{d x} + y) + 2 y \cdot \frac{d y}{d x} = 0 \Rightarrow 2 x + y + x \cdot \frac{d y}{d x} + 2 y \cdot \frac{d y}{d x} = 0 G ro u p \frac{d y}{d x} : x \cdot \frac{d y}{d x} + 2 y \cdot \frac{d y}{d x} = - (2 x + y) \Rightarrow \frac{d y}{d x} = \frac{- ( 2 x + y )}{x + 2 y} A t (1, 2) : \frac{d y}{d x} = \frac{- ( 2 \cdot 1 + 2 )}{1 + 2 \cdot 2} = \frac{- 4}{5} Slope = \frac{- 4}{5}$

6.0Practice Problems on Slope of Tangent

Find the slope of the tangent to $y = ln (x^{2} + 1)$ at x = 1.
Find the slope of the tangent to the curve $x^{2} + y^{2} = 25$ at the point (3, 4).
Find the equation of the tangent line to $y = tan x$ at $x = \frac{π}{6} .$
Parametric: $x = t^{2}, y = t^{3} .$ Find slope at t = 2.
Find the slope of the tangent to the curve $x^{3} + y^{3} = 3 a x y$ at point (a, a).

Frequently Asked Questions

What is called slope of tangent?

Can you find the slope of a tangent line?

What is the slope of this tangent?

Frequently Asked Questions

What is called slope of tangent?

Can you find the slope of a tangent line?

What is the slope of this tangent?

Join ALLEN!

Slope of Tangent

1.0What is the Slope of a Tangent?

2.0What is the Slope of Tangent Formula?

3.0Tangent Line Equation

4.0How to Find the Slope of a Tangent Line?

5.0Solved Examples on Slope of Tangent

6.0Practice Problems on Slope of Tangent

Table of Contents

Join ALLEN!

Slope of Tangent

1.0What is the Slope of a Tangent?

2.0What is the Slope of Tangent Formula?

3.0Tangent Line Equation

4.0How to Find the Slope of a Tangent Line?

5.0Solved Examples on Slope of Tangent

6.0Practice Problems on Slope of Tangent

Table of Contents