Tangent lines Find an equation of the line tangent to the grap...

Question

Tangent lines Find an equation of the line tangent to the graph of f at the given point.

f(x) = sec⁻¹(e^x); (ln 2,π/3)

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Finding the equation of the line tang. Into the graph of F of X is equal to inverse cosicant of the power of 2 X at parentheses the natural log of the square rooted 2 divided by 6. Awesome. So it appears for this particular problem we're ultimately trying to figure out what the equation will be for this line that is tangent to the graph of F of X at the specific point where our point is the natural log of the square to 2 pi divided by 6. So that's our point. Which, as we should recall, is on our, which is an XY. Coordinate set up or a layout, I should say. Awesome. So with that said, our first step in order to solve this problem is we need to define the general equation of the tangent line for any curve, which, as we should recall, we could write this as Y is equal to F of X. And we need to also note that the point of tangency will be X 0 Y 0, so X0Y0. So with that knowledge, we can go ahead and write that. Y 10 where 10 denotes tangents, so Y 10 is equal to F of X0. Multiplied by X minus X0. Plus F of X0. Fantastic. So now our second step is we need to identify the point of tangency, and we need to recall a note that for this particular problem, the point of tangency will be X0Y0 is equal to X0 F of X0. Which is equal to the natural log of the square root of 2, pi divided by 6. Fantastic. So I put the square rooted 2 within parentheses just to remind me that we're saying this is the natural log of the square rooted two. You don't have to write it with parentheses, and you probably will see it without parentheses. So I'll remove the parentheses, but if it helps you denote or remember that you're saying this is the natural log of the square rooted two, then you can keep the parentheses, but they both mean the same thing. So now our next step now for step 3 is we need to adjust the equation from step 1 by substituting in the point of tangency. So we need to state that Y10 is equal to F of the natural log of the square rooted 2 multiplied by X minus the natural log of the square root of 2. Plus F of the natural log of the square root of 2. Fantastic. So now our 4th step is we need to identify the two missing components for the equation as defined in step 3, and we can do this by by starting to evaluate the derivative. So, as we should recall a note, so in order to perform this specific derivative for our function of F of X, we need to recognize that we're gonna be dealing with the chain rule to help us perform this differentiation because E to the power of 2 X is a composite function. So with that said, we need to recall and note that the outer function for this particular prom should be equal to inverse cosicant of you. And the inner function, which we'll call U of X and U of X will be equal to E to the power of 2 X, where once again this E to the power of 2 X is a composite function, and we will also need to recall a note that we will need to apply the chain rule twice as a result of this fact that E to the power of 2 X is a composite function. Awesome. So, with that said, We can go ahead and write, without a doubt that D by DX of inverse cosicant of E to the power of 2 X is equal to D Y. By DU multiplied by DU by DX. Fantastic. So then we can go ahead and write that D by DX of inverse cosicant of E to the power of 2 X. is equal to D by DU of inverse cosicant of U. Multiplied by D by DX of E to the power of 2 X. And this will be all equal to negative 1 divided by the absolute value of U multiplied by the square root of U 2 minus 1, multiplied by E to the power of 2X multiplied by D by DX of. 2X. fantastic. So, this will mean that we can go ahead and write that D by DX of inverse cosicant of E to the power of 2 X. is equal to negative 2 multiplied by the power of 2 X all divided by the absolute value of E to the power of 2 X multiplied by the square root of E of 4 X minus 1. Fantastic. So now, as a result, we can go ahead and write without a doubt that sense E to the power of 2X is greater than 0. We can and that we could say this is the case for all values of X, and because of this fact, we can remove the absolute value, sign, and we can cancel out the common term, and we can go ahead and simply write that D by DX of Inverse cosicant of E to the power of 2 X. is equal to -2 divided by the square root of the power of 4 X minus 1. Fantastic. So now, our next step. Which is for step 5, is we need to evaluate the slope of the tangent line at the point of tangency, which is F of the natural log of the square rooted 2. Fantastic. So when we do that, we will find that F of the natural log of the square rooted 2. is equal to -2 divided by the square root of the power of 4 X minus 1, and we need to note that we're going to be evaluating this at where X is equal to the natural log of the square root of 2. So that will mean that this is equal to -2 divided by the square root of E to the power of 4 multiplied by the natural log of the square root of 2 minus 1. So, when we take this expression and we simplify it down to its most simple form, which I'm gonna skip a few steps, but you can use algebra to rearrange this to simplify. And when you simplify it, you should get the following answer of negative 2 multiplied by the square root of 3 divided by 3, as your final answer for the simplification of this expression. So now our next step for step 6, is we need to evaluate F of the natural log of the square root of 2. And so we're gonna be evaluating this term from step 3, and the term is already given to us in the original point, that is that We could state that F of the natural log of the square rooted 2 is equal to pi divided by 6. So that means for step 7, we could substitute in the value of F of the natural log of the square of 2. So F of the natural log of the squared 2, and F. Of the natural log of the square root of 2, fantastic. And we're gonna be substituting in these two values into our equation from step 3, and when we do that, we will find that Y. 10 is equal to -2 multiplied by the square root of 3 divided by 3, multiplied by X minus the natural log of the square rooted 2 plus pi divided by 6. Fantastic. So what we need to do next for our eighth and final step. is we need to write the equation in its slope intercept form, so meaning we need to take So in order to write our equation in slope intercept form, we need to distribute the terms. So when we do that, we will find that Y10 is equal to -2 multiplied by the square root of 3, divided by 3 multiplied by X, plus 2 multiplied by the square root of 3 divided by 3 multiplied by the natural log. Of the square rooted two, so the natural log of the square rooted 2. Plus pi divided by 6. So what we need to do at this point is we need to simplify this expression to its most simple form, and I'm gonna skip a few steps, but when you do that, you will find that our final answer has to be Y 10. So once again, this is the most simple form of our expression which will state that the tangent line or the Since essentially we're trying to find the equation of the line that is tanged to the graph, that will mean that Y10 is equal to -2 multiplied by the square root of 3. Fantastic, divided by 3, multiplied by X. Plus 1 + 2 multiplied by the square root of 3, multiplied by the natural log of 2. All divided by 6, and that is our final answer. Hooray, we did it. So once again, in conclusion, the equation of the tangent line to the graph of F of X is, so you can write Y 10 or you could just simply write write that Y. is equal to, so you could say that Y is equal to -2 multiplied by the square root of 3 divided by 3 X plus pi plus 2 multiplied by the square of 3, multiplied by the natural lock of 2, all divided by 6. And that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Tangent lines Find an equation of the line tangent to the graph of f at the given point.f(x) = sec−1(ex); (ln 2,π/3)

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Tangent lines Find an equation of the line tangent to the graph of f at the given point.
f(x) = sec⁻¹(e^x); (ln 2,π/3)