Math / Calculus

QuestionThe slope of the tangent line to $y=e^{3 x}$ at $x=-1$ is: $m=3 e^{-3}$ $m=e^{-3}$ $m=3 e^{-1}$ $m=4 e^{2}$

Studdy Solution

STEP 1

What is this asking? We need to find the slope of the tangent line to the function $y = e^{3x}$ at the point where $x = -1$ . Watch out! Remember that the slope of the tangent line is the same as the derivative of the function at that point.
Don't forget the chain rule!

STEP 2

1. Find the derivative.
2. Plug in the value.

STEP 3

We're given the function $y = e^{3x}$ .
Let's define it clearly: $f(x) = e^{3x}$

STEP 4

To find the derivative, we'll use the chain rule.
The chain rule says that the derivative of a composite function (a function inside another function) is the derivative of the outer function, multiplied by the derivative of the inner function.
Here, our outer function is $e^u$ where $u$ represents the inner function, which is $3x$ .

STEP 5

The derivative of $e^u$ with respect to $u$ is simply $e^u$ .
Since $u = 3x$ , this part of the derivative becomes $e^{3x}$ .

STEP 6

The derivative of the inner function, $3x$ , with respect to $x$ is just 3.

STEP 7

Multiplying the results from the outer and inner derivatives, we get the derivative of the whole function: $f'(x) = e^{3x} \cdot 3 = 3e^{3x}$

STEP 8

Now, we want the slope of the tangent at $x = -1$ .
So we substitute $-1$ for $x$ in our derivative: $f'(-1) = 3e^{3 \cdot (-1)}$

STEP 9

This simplifies to: $f'(-1) = 3e^{-3}$

STEP 10

The slope of the tangent line to $y = e^{3x}$ at $x = -1$ is $3e^{-3}$ .
So the correct answer is $m = 3e^{-3}$ .

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QuestionThe slope of the tangent line to $y=e^{3 x}$ at $x=-1$ is: $m=3 e^{-3}$ $m=e^{-3}$ $m=3 e^{-1}$ $m=4 e^{2}$

Studdy Solution

STEP 1

What is this asking? We need to find the slope of the tangent line to the function $y = e^{3x}$ at the point where $x = -1$ . Watch out! Remember that the slope of the tangent line is the same as the derivative of the function at that point.
Don't forget the chain rule!

STEP 2

1. Find the derivative.
2. Plug in the value.

STEP 3

We're given the function $y = e^{3x}$ .
Let's define it clearly: $f(x) = e^{3x}$

STEP 4

STEP 5

The derivative of $e^u$ with respect to $u$ is simply $e^u$ .
Since $u = 3x$ , this part of the derivative becomes $e^{3x}$ .

STEP 6

The derivative of the inner function, $3x$ , with respect to $x$ is just 3.

STEP 7

Multiplying the results from the outer and inner derivatives, we get the derivative of the whole function: $f'(x) = e^{3x} \cdot 3 = 3e^{3x}$

STEP 8

Now, we want the slope of the tangent at $x = -1$ .
So we substitute $-1$ for $x$ in our derivative: $f'(-1) = 3e^{3 \cdot (-1)}$

STEP 9

This simplifies to: $f'(-1) = 3e^{-3}$

STEP 10

The slope of the tangent line to $y = e^{3x}$ at $x = -1$ is $3e^{-3}$ .
So the correct answer is $m = 3e^{-3}$ .

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QuestionThe slope of the tangent line to y=e3xy=e^{3 x}y=e3x at x=−1x=-1x=−1 is: m=3e−3m=3 e^{-3}m=3e−3 m=e−3m=e^{-3}m=e−3 m=3e−1m=3 e^{-1}m=3e−1 m=4e2m=4 e^{2}m=4e2

Studdy Solution

STEP 1

What is this asking? We need to find the slope of the tangent line to the function y=e3xy = e^{3x}y=e3x at the point where x=−1x = -1x=−1. Watch out! Remember that the slope of the tangent line is the same as the *derivative* of the function at that point.Don't forget the chain rule!

STEP 2

1. Find the derivative.2. Plug in the value.

STEP 3

We're given the function y=e3xy = e^{3x}y=e3x.Let's **define** it clearly: f(x)=e3xf(x) = e^{3x}f(x)=e3x

STEP 4

STEP 5

The derivative of eue^ueu with respect to uuu is simply eue^ueu.Since u=3xu = 3xu=3x, this part of the derivative becomes e3xe^{3x}e3x.

STEP 6

The derivative of the inner function, 3x3x3x, with respect to xxx is just **3**.

STEP 7

Multiplying the results from the outer and inner derivatives, we get the **derivative** of the whole function: f′(x)=e3x⋅3=3e3xf'(x) = e^{3x} \cdot 3 = 3e^{3x}f′(x)=e3x⋅3=3e3x

STEP 8

Now, we want the slope of the tangent at x=−1x = -1x=−1.So we **substitute** −1-1−1 for xxx in our derivative: f′(−1)=3e3⋅(−1)f'(-1) = 3e^{3 \cdot (-1)}f′(−1)=3e3⋅(−1)

STEP 9

This simplifies to: f′(−1)=3e−3f'(-1) = 3e^{-3}f′(−1)=3e−3

STEP 10

The slope of the tangent line to y=e3xy = e^{3x}y=e3x at x=−1x = -1x=−1 is 3e−33e^{-3}3e−3.So the **correct answer** is m=3e−3m = 3e^{-3}m=3e−3.

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QuestionThe slope of the tangent line to $y=e^{3 x}$ at $x=-1$ is: $m=3 e^{-3}$ $m=e^{-3}$ $m=3 e^{-1}$ $m=4 e^{2}$

What is this asking? We need to find the slope of the tangent line to the function $y = e^{3x}$ at the point where $x = -1$ . Watch out! Remember that the slope of the tangent line is the same as the derivative of the function at that point.
Don't forget the chain rule!

1. Find the derivative.
2. Plug in the value.

We're given the function $y = e^{3x}$ .
Let's define it clearly: $f(x) = e^{3x}$

The derivative of $e^u$ with respect to $u$ is simply $e^u$ .
Since $u = 3x$ , this part of the derivative becomes $e^{3x}$ .

The derivative of the inner function, $3x$ , with respect to $x$ is just 3.

Multiplying the results from the outer and inner derivatives, we get the derivative of the whole function: $f'(x) = e^{3x} \cdot 3 = 3e^{3x}$

Now, we want the slope of the tangent at $x = -1$ .
So we substitute $-1$ for $x$ in our derivative: $f'(-1) = 3e^{3 \cdot (-1)}$

This simplifies to: $f'(-1) = 3e^{-3}$

The slope of the tangent line to $y = e^{3x}$ at $x = -1$ is $3e^{-3}$ .
So the correct answer is $m = 3e^{-3}$ .