Slope of a curve at a point - AP Calculus AB

The function is not differentiable at . So the derivative at does not exist.

Find the derivative of the curve using the power rule.

In mathematical terms, the power rule states,

Therefore the derivative is,

Next, plug in the x-value to find the slope of the curve at , which gives you a final answer of

Expand the binomial and combine like terms to get,

Next, take the derivative of the polynomial using the power rule, which is in mathematical terms,

Therefore,

Lastly, plug in for , to get the slope at the point.

This gives you your final answer of

To find the slope at a point, we take the derivative of , substitute in , and simplify.

.

.

First, find the derivative. You should get .

Next, plug in x=1.

You should get .

First, use the chain rule to find f'(t).

You should get .

Next, plug in t=0 to get f'(0)=.

Derivatives are slope finders. Therefore to find the slope at the given point, we need to find the derivative of the 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.

So, we get

From there we plug in our x value.

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

To find the derivative we need to use product rule. Product rule states that we take the derivative of the first function and multiply it by the derivative of the second function and then add that with the derivative of the second function multiplied by the given first function. To find the derivative of each separate function we need to use 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

Use power rule and we get :

From here, to find the slope at the given point we plug in "2" for x.

)(-3(2%5E2))&space;+&space;(-&space;(2%5E3)+4)(2(2)+2) $"(2^2$ + $2(2))(-3(2^2$)) + (- $(2^3$)+4)(2(2)+2)")

This comes out to equal

Find the rate of change of f(x) at the point (4,12)

We are asked to find a rate of change, so begin by finding the first derivative.

Now, we need the rate of change at (4,12). What matters most is the x value, simply plug it into our derivative and solve for y

So, our answer is 1271

To find the slope at a point of our function, we need to find its derivative first.

Using the product rule (and the chain rule within this product rule application), we have

.

Plugging the -value of our point into this equation, we get our desired slope of

.

Find the slope of the line tangent to f(x) at the point x=0

To find the slope of a tangent line, we must first find the derivative of our function.

Let's recall a few rules to help us out.

1. The derivative of a monomial can be found by multiplying the coefficient by the exponent, and then decreasing the exponent by 1.

1. The derivative of e to the x is e to the x

1. The derivative of cosine is negative sine

Now, put it all together to get:

Lastly, we need to plug in 0 for x and solve our equation.

So, our answer is 0

To find the slope at a point, we first find the derivative of our function, and then substitute in the -value of our point.

, so , and plugging in our -value gives .