find the derivative

ln [x^3 +((x+3)^3)((x^2)+4)^7

You left out a ] somewhere

Hmm the logical fix would be

ln [x^3 +((x+3)^3)]((x^2)+4)^7

a^3 + b^3 = (a+b)(a^2-ab+b^2), so

x^3 + (x+3)^3 = (x+(x+3))(x^2 - x(x+3) + (x+3)^2)
= (2x+3)(x^2 + 3x + 9)

and the log of the product then becomes

y = ln(2x+3) + ln(x^2+3x+9) + 7ln(x^2+4)

y' = 2/(2x+3) + (2x+3)/(x^2+3x+9) + 14x/(x^2+4)

If my placement of [] is wrong, please feel free to clarify

Find the derivative of the function

y=ln(2x/x+3)

To find the derivative of the given expression, we can use the chain rule and product rule. Let's break down the steps:

Step 1: Determine the overall function and identify the parts to differentiate separately.

The given expression is ln [x^3 +((x+3)^3)((x^2)+4)^7]. Here, we have an outer logarithmic function and an inner expression.

Step 2: Apply the chain rule.

The derivative of the outer logarithmic function ln(u) is (1/u) * du/dx.

Step 3: Differentiate the inner expression.

For differentiating the inner expression, we'll apply the product rule as it contains two separate terms.

The first term is x^3, and its derivative is 3x^2.

The second term is ((x+3)^3)((x^2)+4)^7. Let's break it down further to identify more manageable parts:

Let's denote u = (x+3)^3 and v = (x^2 + 4)^7.

Now, we can find the derivatives of u and v:

For u:
Using the chain rule, derivative of u = 3(x+3)^2 * (d(x+3)/dx).

For v:
Using the chain rule, derivative of v = 7(x^2 + 4)^6 * (d(x^2 + 4)/dx).

Step 4: Find the overall derivative using the chain rule and product rule.

Now that we have all the necessary derivatives, we can calculate the overall derivative:

d/dx [ln [x^3 +((x+3)^3)((x^2)+4)^7]]
= (1/u) * du/dx

The value of u is (x+3)^3, so we substitute its derivative:

= (1/(x+3)^3) * 3(x+3)^2 * (d(x+3)/dx)

Next, the value of v is (x^2 + 4)^7, and we substitute its derivative:

= (1/(x+3)^3) * 3(x+3)^2 * (d(x+3)/dx) + v * 7(x^2 + 4)^6 * (d(x^2 + 4)/dx)

Finally, we can simplify and combine like terms to obtain the derivative of the given expression.