Question

Pешение производных Y=(x ^6+x ^5-5/x) ^1/3 Y= sin^3*5x+e^7x Y=tg^3*4x

Answer 1

Рроьмпщпшшмжмрдлилшсопвллралрралн

Answer 2

To find the derivatives of the given functions, we'll use basic rules of differentiation. Let's calculate the derivatives step by step:

1. For Y = (x^6 + x^5 - 5/x)^(1/3):

We'll use the chain rule, which states that if we have a function of the form (u)^n, then its derivative is n * u^(n-1) * u'.

u = x^6 + x^5 - 5/x n = 1/3

Y' = (1/3) * (x^6 + x^5 - 5/x)^((1/3) - 1) * (x^6 + x^5 - 5/x)'

Now, let's find (x^6 + x^5 - 5/x)':

(x^6 + x^5 - 5/x)' = (6x^5 + 5x^4 + 5/x^2)

Putting it back into Y':

Y' = (1/3) * (x^6 + x^5 - 5/x)^((1/3) - 1) * (6x^5 + 5x^4 + 5/x^2)

Simplifying:

Y' = (1/3) * (x^6 + x^5 - 5/x)^(-2/3) * (6x^5 + 5x^4 + 5/x^2)

2. For Y = sin^3(5x) + e^(7x):

a) Derivative of sin^3(5x): Using the chain rule and the derivative of sin(x):

d/dx [sin^3(5x)] = 3 * sin^2(5x) * cos(5x) * 5 = 15 * sin^2(5x) * cos(5x)

b) Derivative of e^(7x): Using the chain rule and the derivative of e^x:

d/dx [e^(7x)] = 7 * e^(7x)

Now, the derivative of Y with respect to x (Y') is:

Y' = 15 * sin^2(5x) * cos(5x) + 7 * e^(7x)

3. For Y = tan^3(4x):

Using the chain rule and the derivative of tan(x):

d/dx [tan^3(4x)] = 3 * tan^2(4x) * sec^2(4x) * 4 = 12 * tan^2(4x) * sec^2(4x)

Therefore, the derivative of Y with respect to x (Y') is:

Y' = 12 * tan^2(4x) * sec^2(4x)

Please note that the trigonometric functions' derivatives are dependent on the chain rule and the derivative of their corresponding trigonometric functions (e.g., sin(x), cos(x), tan(x), etc.). Also, ensure that you understand the concept of the chain rule well, as it's crucial for finding derivatives of composite functions like these.