Derivatives and tangent lines
b. Determine an equation o...

Question

Derivatives and tangent lines

b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.

f(x) = √3x; a= 12

Accepted Answer

Welcome back, everyone, to another video. Calculate the equation of the attention line to the function P X equals square root of 5 X at the point A P A where A is equal to 25. We're given 4 answer choices. A says Y equals square root of 5 divided by 10. X minus 52 5 divided by 2. By equals square root of 10 multiplied by x plus 52 5 divided by 2 C Y equals +25 divided by 10 X plus 5 divided by 2 and D Y equals 25 divided by 10 X plus 5 square root of 5. So if we're looking for the equation of the tent plan, we have to recall that it has a form of Y equals F at point X not multiplied by X minus X plus F at point X0. In this case, we are given a function p of X, right? So we can say that the general expression is the derivative of p at point X0. In this case, X0 is simply a. That's the point of interest multiplied by X minus A plus the value of the function at point A. Now let's not this that we Want to identify P of A, which is, which means p of 25, then our A is 25, and we also need the derivative of P at 0.25. All in all, we just need to identify two terms. Let's begin with P of A, and this means that we are simply substituting 25 into the function. So P of 25 equals square root of 5 multiplied by 25. And now we can simplify that. This is square root of 5 multiplied by square root of 25, which gives us 5 square root of 5. And now we need to identify the derivative in its general form P X. We can see that by factoring out the constant square root of 5 and identifying the derivative of X to the power of 1/2, which is square root of X. And we're going to apply the power rules so factoring out the constant, we get root of 5. Applying the power rule, we bring down the constant, that's 1/2. And the original exponent has to decrease by 1 unit, so 1/2 minus 1 gives us negative 1/2. And now this simply gives us square root of 5, and then we have a denominator which is 2. Since X has a negative exponent, we can bring it down to the denominator, and X to the power of 1/2 gives us square root of X. Now, we are not going to rationalize it because we're looking for the derivative of P at 25, so we can just substitute. 25 for x and we get square root of 5 divided by 2 multiplied by square root of 25, which gives us square root of 5 divided by 2 multiplied by 5, and that's 10. Well done. So we have all of the terms that we need, and now we can write the equation of the tangent line starting with the derivative of p at 25. We substitute our slope square root of 5 divided by 10 multiplied by X minus 25, and then we are adding P at 25, which is 5 square root of 5. And now if we distribute the terms, we get square root of 5 divided by 10 X minus 25 divided by 10. In the simplified form, we can divide 25 by 5. This gives us 5, and in the denominator, we're going to get 2, right? So 5 divided by 2 square root of 5. But instead of performing this, well, we can essentially keep the same denominator if we want. However, we can write it in a simplified form, right? 25 divided by 10 is exactly the same as 5 divided by 2. And then we're adding 5 square roots of 5. Now from here, well, we can essentially take the common denominator to combine the constants, and we're going to get square root of 5 divided by 10 X minus 5 square root of 5 divided by 2 plus. Now we want to obtain a denominator of 2 for the for the final term, and this means that we're multiplying 5 by 2, right? So this gives us 10 square root of 5. And now what we want to do is combine the like terms, which gives us our final expression, square root of 5 divided by 10, multiplied by X, and now -5 plus 10 gives us 5. Followed by a square root of 5 divided by 2. Now let's consider the answer choices. If we carefully analyze the answer choices, we can conclude that the correct answer corresponds to the choice B. Thank you for watching.

Derivatives and tangent lines
b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.
f(x) = √3x; a= 12

Key Concepts

Derivative

Tangent Line

Point-Slope Form

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