# AP Calculus AB : Tangent line to a curve at a point and local linear approximation

## Example Questions

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### Example Question #1 : Derivatives

Differentiate,

Explanation:

Differentiate,

Strategy

This one at first glance appears difficult even if we recognize that the chain rule is needed; we have a function within a function within a function within a function. To avoid making mistakes, it's best to start by defining variables to make the calculation easier to follow.

Let's start with the outermost function, we will write  as a function of  by setting,

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Similarly, define  to write  as a function of

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Write  as a function of

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Finally, define the inner-most function, , as the function of

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Since  we will just substitute that in and move to the front.

That was easy enough, now just write everything in terms of  by going back to the definitions of   and

### Example Question #1 : Tangent Line To A Curve At A Point And Local Linear Approximation

Find the tangent line. Given the point (1,2)

Explanation:

To find the tangent line at the given point, we need to first take the derivative of the given function.

Power Rule:

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent.

Therefore,  becomes

From there we plug in the "1" from the point to get our m value of the equation . When we plug in "1" to y' we get m=-1. Then from there, we will plug our point into  now that we have found m to find our b value. So,

Therefore, the tangent line is equal to

### Example Question #3 : Derivatives

Find the line tangent at the point (0,1)

Explanation:

To find the tangent line at the given point, we need to first take the derivative of the given function. The rule for functions with "e" in it says that the derivative of  However with this function there is also a 3 in the exponent so we will also use chain rule. Chain Rules states that we work from the outside to the inside. Meaning we will take the derivative of the outside of the equation and multiply it by the derivative of the inside of the equation.

To put this into equation it will look like

From there we plug in the "0" from the point to get our m value of the equation . When we plug in "0" to y' we get m=3. Then from there, we will plug our point into   now that we have found m to find our m value. So,

then plug this all back into the equation once more and we are left with

### Example Question #4 : Derivatives

Find the tangent line given the point (2,4) and the equation

Explanation:

To find the tangent line at the given point, we need to first take the derivative of the given function using Power Rule

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent.

From there we plug in the "2" from the point to get our m value of the equation . When we plug in "2" to y' we get m=8. Then from there, we will plug our point into  now that we have found m to find our b value. So,

Plug this back into

### Example Question #1 : Derivative At A Point

Find the equation of the line tangent to the curve at the point where

Explanation:

Find the equation of the line tangent to the curve  at the given point

The slope of the line tangent at the given point will be equal to the derivative of  at that point. Compute the derivative and find the slope for our line:

Evaluate the secant term:

Therefore slope of the tangent line is simply:

So now we know the slope of the tangent line and can write the equation then solve for

In order to solve for  we need one point on the line. Use the point where the tangent line meets the curve. Use the original function to find the "y" coordinate at this point:

We now have our point:

Use the point to find

### Example Question #6 : Derivatives

Find the slope of the line tangent to the curve of d(g) when g=6.

Explanation:

Find the slope of the line tangent to the curve of d(g) when g=6.

All we need here is the power rule. This states that to find the derivative of a polynomial, simply subtract 1 from each exponent and then multiply each term by their original exponent.

Constant terms will drop out when we do this, and linear terms will become constants.

From here substitute in g=6.

### Example Question #7 : Derivatives

Give the equation of the line tangent to the graph of the equation

at the point .

Explanation:

The tangent line to the graph of  at point  is the line with slope  that passes through that point. Find the derivative :

Apply the sum rule:

The tangent line is therefore the line with slope 5 through .Apply the point-slope formula:

### Example Question #1 : Derivatives

Give the equation of the line tangent to the graph of the equation

at the point .

None of the other choices gives the correct response.

None of the other choices gives the correct response.

Explanation:

The tangent line to the graph of  at point  is the line with slope  that passes through that point. Find the derivative :

Apply the constant multiple and sum rules:

Set  and  and apply the chain rule.

Substituting back:

Evaluate  using substitution:

The tangent line is therefore the line with slope  through  is a -intercept, so apply the slope-intercept formula to get the equation

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This is not among the choices given.

### Example Question #9 : Derivatives

Find the equation of the line parallel to the function  at , and passes through the point

Explanation:

We first start by finding the slope of the line in question, which we do by taking the derivative of  and evaluate at

We then use point slope form to get the equation of the line at the point

### Example Question #10 : Derivatives

Find the equation of the line tangent to  at the point .