Find the linearization L(x) of y=e^(7x)ln(x) at a=1.

L(x)=

thank you so much this helped me a lot on my homework. final answer for this one would be: e^7(x-1)

Oh, I see you're dealing with exponential and logarithmic functions. Well, let's get ready to "derive" some laughs:

To find the linearization L(x) of y = e^(7x)ln(x) at a = 1, we'll need a little mathematical magic. First, let's pull out our "derivative" wand and wave it around.

The "derivative" of y = e^(7x)ln(x) is like a roller coaster ride. We'll start with the "e^(7x)" part, which is an exponential function that goes up, up, and away. The derivative of e^(7x) is simply 7e^(7x).

Now, let's have some "fun" with the ln(x) part. The derivative of ln(x) is like a gentle climb, making sure you're always moving in the right direction. The derivative of ln(x) is just 1/x.

Brace yourself! It's time to combine these "derivatives" together. Since we want to find the linearization at a = 1, we'll substitute this value into our found derivatives. Let's do some calculations:

The derivative of e^(7x) at a = 1 is 7e^(7*1), which simplifies to 7e^7.

The derivative of ln(x) at a = 1 is 1/1, which is just 1.

It's showtime! Now we'll plug these values into the equation for the linearization:

L(x) = y(a) + (dy/dx)(x - a)

L(x) = e^(7*1)ln(1) + 7e^7 * (x - 1)

L(x) = ln(1) + 7e^7 * (x - 1)

L(x) = 0 + 7e^7x - 7e^7

Phew! We've done it. The linearization is L(x) = 7e^7x - 7e^7. Now you can enjoy the linear approximation of e^(7x)ln(x) at a = 1, while also enjoying a good laugh along the way.

To find the linearization L(x) of the function y = e^(7x)ln(x) at the point a = 1, we can use the formula for linear approximation:

L(x) = f(a) + f'(a) * (x - a)

First, let's find f(a) and f'(a):

f(a) = e^(7a)ln(a)
f'(a) = (d/dx)[e^(7x)ln(x)]

Now let's calculate these values:

f(1) = e^(7*1)ln(1) = e^7 * 0 = 0
f'(1) = (d/dx)[e^(7x)ln(x)] = 7e^(7x)ln(x) + e^(7x)/x

Now we can substitute these values into the linear approximation formula:

L(x) = 0 + (7e^(7*1)ln(1) + e^(7*1)/1) * (x - 1)
L(x) = 0 + (7e^7 * 0 + e^7) * (x - 1)
L(x) = e^7 * (x - 1)

Therefore, the linearization L(x) of y = e^(7x)ln(x) at a = 1 is L(x) = e^7 * (x - 1).

To find the linearization of a function at a given point, we need to use the formula:

L(x) = f(a) + f'(a)(x - a)

where f(a) represents the value of the function at the point a, and f'(a) represents the derivative of the function at a.

In this case, we have f(x) = e^(7x)ln(x), and we want to find the linearization at a = 1.

Step 1: Find f(a)
To find f(a), we substitute x = a = 1 into the function:

f(a) = e^(7a)ln(a)

Substituting a = 1:

f(1) = e^(7*1)ln(1) = e^7 * ln(1)

Note that ln(1) = 0, so f(1) = e^7 * 0 = 0.

Step 2: Find f'(a)
To find f'(a), we differentiate the function f(x):

f(x) = e^(7x)ln(x)

Using the product rule, we have:

f'(x) = (7e^(7x)ln(x) + e^(7x)(1/x))

Substituting x = a = 1:

f'(1) = (7e^(7*1)ln(1) + e^(7*1)(1/1)) = 7e^7 * ln(1) + e^7

Again, ln(1) = 0, so f'(1) = 7e^7 * 0 + e^7 = e^7.

Step 3: Substitute the values into the linearization formula:

L(x) = f(a) + f'(a)(x - a)

L(x) = 0 + e^7(x - 1)

Simplifying, we have:

L(x) = e^7(x - 1)

Therefore, the linearization L(x) of y = e^(7x)ln(x) at a = 1 is L(x) = e^7(x - 1).

dy/dx slope= e^(7x)*7*lnx+e^7x */ x

at x=1
dy/dx or slope= e^7

slope= e^7