Question

Найти производную y=ln (8x-3/5x^2+4)! Помогите пожалуйста!!!

Answer 1

$y=ln \frac{8x-3}{5x^2+4}\\\\y'=\frac{5x^2+4}{8x-3}\cdot \frac{8(5x^2+4)-10x(8x-3)}{(5x^2+4)^2}=\frac{40x^2+32-80x^2+30x}{(8x-3)(5x^2+4)}=\\\\= \frac{-40x^2+30x+32}{(8x-3)(5x^2+4)}$

Answer 2

Calculating the Derivative of y = ln((8x - 3)/(5x^2 + 4))

To find the derivative of the function y = ln((8x - 3)/(5x^2 + 4)), we can use the quotient rule. The quotient rule states that if we have a function of the form f(x)/g(x), then the derivative is given by:

dy/dx = (g(x) * f'(x) - f(x) * g'(x)) / (g(x))^2

In this case, f(x) = ln((8x - 3)/(5x^2 + 4)), and g(x) = 1. To find the derivative, we need to find f'(x) and g'(x).

Let's start by finding f'(x), the derivative of ln((8x - 3)/(5x^2 + 4)). To do this, we can use the chain rule, which states that if we have a function of the form ln(u(x)), then the derivative is given by:

d/dx ln(u(x)) = (1/u(x)) * u'(x)

In this case, u(x) = (8x - 3)/(5x^2 + 4). Let's find u'(x) first.

To find u'(x), we can use the quotient rule again. The quotient rule states that if we have a function of the form f(x)/g(x), then the derivative is given by:

du/dx = (g(x) * f'(x) - f(x) * g'(x)) / (g(x))^2

In this case, f(x) = 8x - 3, and g(x) = 5x^2 + 4. Let's find f'(x) and g'(x).

f'(x) = 8 g'(x) = 10x

Now, let's substitute these values into the formula for u'(x):

u'(x) = ((5x^2 + 4) * 8 - (8x - 3) * 10x) / (5x^2 + 4)^2

Simplifying the expression, we get:

u'(x) = (40x^2 + 32 - 80x^2 + 30x) / (5x^2 + 4)^2

u'(x) = (-40x^2 + 30x + 32) / (5x^2 + 4)^2

Now that we have u'(x), we can find f'(x) by substituting u'(x) into the formula for f'(x):

f'(x) = (1/u(x)) * u'(x)

f'(x) = (1/((8x - 3)/(5x^2 + 4))) * (-40x^2 + 30x + 32) / (5x^2 + 4)^2

Simplifying the expression, we get:

f'(x) = (-40x^2 + 30x + 32) / (8x - 3)^2

Now that we have f'(x) and g'(x), we can substitute these values into the formula for dy/dx:

dy/dx = (g(x) * f'(x) - f(x) * g'(x)) / (g(x))^2

dy/dx = (1 * (-40x^2 + 30x + 32) - ln((8x - 3)/(5x^2 + 4)) * 10x) / (1)^2

Simplifying the expression, we get:

dy/dx = (-40x^2 + 30x + 32 - 10x * ln((8x - 3)/(5x^2 + 4))) / (8x - 3)^2

Therefore, the derivative of y = ln((8x - 3)/(5x^2 + 4)) is:

dy/dx = (-40x^2 + 30x + 32 - 10x * ln((8x - 3)/(5x^2 + 4))) / (8x - 3)^2