Question

Помогите, пожалуйста найти производную первого порядка (ln(sin(x/((x^4)-3*x))))

Answer 1

(ln(sin( x/ (x^4 -3x)) )'=1/ sin (x/(x^4 -3x)) * (sin(x/(x^4 -3x) )'=
=1/sin(x/(x^4 -3x) ) * cos(x/(x^4 -3x) * (x/(x^4 -3x) )'=
=(cos(x/(x^4 -3x) ) /sin(x/(x^4 -3x))  * (1*(x^4 -3x)-x(4x^3-3) )/(x^4 -3x)^2=
=ctg (x/ (x^4 -3x) )  *(x^4 -3x  -4x^4+3x) /(x^4 -3x)^2=
=(-3x^4  *ctg(x/(x^4 -3x) ) ) /(x^4-3x)^2 

Answer 2

Finding the First-Order Derivative

To find the first-order derivative of the function ln(sin(x/((x^4)-3*x))), we can use the chain rule and the derivative of the natural logarithm function. The chain rule states that if we have a composite function, f(g(x)), then the derivative is f'(g(x)) * g'(x).

Let's calculate the first-order derivative step by step.

Calculating the Derivative

The function is ln(sin(x/((x^4)-3*x))).

Using the chain rule, the derivative of ln(u) is u'/u, where u is the inner function.

The derivative of sin(u) is cos(u) * u'.

Therefore, the first-order derivative of the given function is: y' = (cos(x/((x^4)-3*x)) * ((x^4)-3*x))' / (sin(x/((x^4)-3*x))

Now, let's cite the relevant search snippet for further clarification. ``` y'=(5tg4x)'logsinx+(5tg4x)(lnsinx/ln3)'-[((x-4x+5))'(3^cos5x)-(3^cos5x)'(x-4x+5)]/(3^cos5x)^2=(20/cos5x)logsinx+(5tg4x)(cosx/sinx)/... ``` The cited snippet provides a derivative calculation using a similar approach, demonstrating the application of the chain rule and the derivative of the natural logarithm function.

Conclusion

The first-order derivative of ln(sin(x/((x^4)-3*x))) is given by the expression mentioned above.