Find an equation for the tangent line to the graph at the specified value of x

Question

1 - y = xcos3x; x = π

 

2 - y = sec ³ (π/2 - x); x = -π/2

 

3 - y = Tan(4x ²  ); x = √π

 

 

Hey guys please help me with these ones, i am studying for finals, if possible try to solve them step by step in detail so i can understand... 

 

Answer 1

In order to find a tangent line, we need to take a derivative of the function. Could you do that? If not, you could use one of the derivative calculators online.

 

for example:

 

(x*cos(3*x))'

 

using Product Rule for Derivatives

(x)'*cos(3*x)+x*(cos(3*x))'

1*cos(3*x)+x*(cos(3*x))'

 

chain rule for cos(3*x)

1*cos(3*x)+x*-sin(3*x)*(3*x)'

1*cos(3*x)+x*-sin(3*x)*((3)'*x+3*(x)')

1*cos(3*x)+x*-sin(3*x)*(0*x+3*(x)')

1*cos(3*x)+x*-sin(3*x)*(0*x+3*1)

1*cos(3*x)+x*3*(-sin(3*x))

1*cos(3*x)+x*-3*sin(3*x)

cos(3*x)-(3*x*sin(3*x))

 

Now we need to put pi in place of x and calculate. 

 

cos(3*pi) = -1.

 

sin (3*pi) = 0, that means that (3*x*sin(3*x)) = 0.

 

So, in conclusion, to find the tangent line at a point x0, you need to:

1. Take the derivative of your function.

2. Plug in your x-value, x0, into the derivative to get your slope.
If the derivative ends up just being a constant, like -1 in our situation, then just use that as your slope.

 

Was it helpful?

Do you need me to go over other problems?

 

 

 

 

 

Answer 2

Hi Eric

To find the tangent line you need two pieces of information:

(A) The point (x₀, y₀)on the graph

(B) The slope of the graph at that point

 

To find (A) you plug the x-value they give you into the original function. To get (B)you take the derivative and plug the x-value into that.

 

So for question 1:

(A) Plug π into the function: y = π*cos3π = -π (hopefully you can find this without a calculator - use the unit circle to find trig values)

So your coordinates are (π, -π).

 

(B) Take the derivative. You will need the product rule here.

y′ = cos3x - 3xsin3x

Evaluate at π:

y′(π) = cos3π - 3πsin3π = -1. This is your slope.

 

So now you can use point-slope form to find the line.Point slope form is y - y₀ = m(x - x₀)

We get y - (-π) = -1(x - π)

Simplify: y + π = -1(x - π)

You could leave it like this but better is to solve for y.

We have y + π = -x + π

Or: y = -x. This is the answer to question 1.

 

The other two are similar but a little harder since you have to use the chain rule. For instance for the second question, the function is

y = sec³(π/2 - x)

So the derivative will be

y′ = 3sec²(π/2 - x)*sec(π/2 - x)tan(π/2 - x)*(-1)

    = -3sec³(π/2 - x)tan(π/2 - x)