For the function below, find the slope of the tangent line at x=2 (5x^(4)+4)/(-2x^(3)+7)
Can someone explain how to do this?
To find the slope of the tangent line at a specific point, we can use the derivative of the function. The derivative measures how the function changes as x varies.
To find the derivative of the given function, we can use the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x) / h(x), then the derivative of f(x) is given by [h(x) * g'(x) - g(x) * h'(x)] / [h(x)]^2.
Applying the quotient rule to the given function, f(x) = (5x^4 + 4) / (-2x^3 + 7), we have g(x) = 5x^4 + 4 and h(x) = -2x^3 + 7.
Now, let's find the derivatives of g(x) and h(x):
To find the derivative of g(x), we differentiate each term:
g'(x) = 20x^3 + 0 = 20x^3
To find the derivative of h(x), we differentiate each term:
h'(x) = -2(3x^2) + 0 = -6x^2
Now, substitute these values into the derivative formula:
f'(x) = [h(x) * g'(x) - g(x) * h'(x)] / [h(x)]^2
= [(-2x^3 + 7)(20x^3) - (5x^4 + 4)(-6x^2)] / [(-2x^3 + 7)^2]
Simplifying the expression, we get:
f'(x) = [(-40x^6 + 140x^3) - (-30x^6 - 24x^2)] / [(-2x^3 + 7)^2]
= [(-40x^6 + 140x^3 + 30x^6 + 24x^2)] / [(-2x^3 + 7)^2]
= [-10x^6 + 140x^3 + 24x^2] / [(-2x^3 + 7)^2]
This is the derivative function. To find the slope of the tangent line at x = 2, we need to evaluate the derivative at x = 2.
Substitute x = 2 into the derivative function:
f'(2) = [-10(2^6) + 140(2^3) + 24(2^2)] / [(-2(2^3) + 7)^2]
= [-10(64) + 140(8) + 24(4)] / [(-16 + 7)^2]
= [-640 + 1120 + 96] / [(-9)^2]
= [576] / [81]
= 7.11 (approximately)
Therefore, the slope of the tangent line at x = 2 is approximately 7.11.