1. find the equation of the tangent line at x=1 to y=f(x) where f(x)=(3x^2)/(5x^2+7x)

Can someone help walk me through the steps of this problem. Do I start with finding the derivative of the function using the quotient rule?

Yes, you are correct. To find the equation of the tangent line at x=1 to the function y=f(x), you need to find the derivative of the function using the quotient rule.

Let's start by finding the derivative of f(x). The quotient rule states that if you have a function f(x) = g(x)/h(x), then the derivative of f(x) is given by:

f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2

In our case, g(x) = 3x^2 and h(x) = 5x^2 + 7x. Let's differentiate g(x) and h(x) separately:

g'(x) = d/dx (3x^2) = 6x

h'(x) = d/dx (5x^2 + 7x) = 10x + 7

Now we can substitute these values into the quotient rule to find f'(x):

f'(x) = (6x * (5x^2 + 7x) - 3x^2 * (10x + 7)) / ((5x^2 + 7x)^2)

Simplifying this equation further:

f'(x) = (30x^3 + 42x^2 - 30x^3 - 21x^2) / ((5x^2 + 7x)^2)

f'(x) = (21x^2) / ((5x^2 + 7x)^2)

Now that we have the derivative f'(x), we can find the slope of the tangent line by plugging in the value x=1:

slope = f'(1) = (21(1)^2) / ((5(1)^2 + 7(1))^2)

slope = 21 / (5 + 7)^2 = 21/12^2 = 21/144

Therefore, the slope of the tangent line at x=1 is 21/144.

To find the equation of the tangent line, we need both the slope and a point on the line. We already know the x-coordinate of the point, which is x=1. To find the corresponding y-coordinate, we can substitute x=1 back into the original function f(x):

f(1) = (3(1)^2) / (5(1)^2 + 7(1)) = 3/12 = 1/4

Therefore, the y-coordinate of the point is y=1/4.

Now we can use the point-slope form of a line to find the equation:

y - y1 = m(x - x1)

where m is the slope and (x1, y1) is the point on the line.

Plugging in the values we have:

y - 1/4 = (21/144)(x - 1)

Multiplying both sides by 144 to eliminate the fraction:

144y - 36 = 21(x - 1)

Rearranging the equation:

21x - 144y = 165

This is the equation of the tangent line at x=1 to the function y=f(x).