Question

Найти производную: 1) f(x)=tg(x+pi/4) 2) f(x)=sin2x+tgx 3) f(x)=4-1/4tgx 4) f(x)=x^2ctgx

Answer 1

Ответ:

1.

$f'(x) = \frac{1}{ { \cos }^{2} (x + \frac{\pi}{4}) }$

2.

$f'(x) = 2 \cos(2x) + \frac{1}{ { \cos}^{2} x}$

3.

$f'(x) = (4 - \frac{1}{4tgx} ) '= - \frac{1}{4} \times ( - {tg}^{ - 2} x) \times \frac{1}{ { \cos }^{2}x } = \\ = \frac{1}{4} \times \frac{ { \cos }^{2}x }{ { \sin }^{2}x } \times \frac{1}{ { \cos }^{2} x} = \frac{1}{4 { \sin }^{2} x}$

4.

$f'(x) = ( {x}^{2} )'ctgx + (ctgx) '\times {x}^{2} = \\ = 2xctgx - \frac{ {x}^{2} }{ { \sin }^{2} x}$

Answer 2

1) f(x) = tg(x+pi/4)

To find the derivative of f(x) = tg(x+pi/4), we can use the chain rule. The chain rule states that if we have a composite function, f(g(x)), then the derivative is given by f'(g(x)) * g'(x).

Let's break down the function:

f(x) = tg(x+pi/4)

Here, g(x) = x + pi/4, and f(x) = tg(g(x)).

Now, we can find the derivative step by step:

Step 1: Find the derivative of the inner function g(x). g'(x) = 1

Step 2: Find the derivative of f(g(x)). f'(u) = sec^2(u), where u = g(x)

Step 3: Substitute u with g(x). f'(g(x)) = sec^2(g(x))

Step 4: Multiply f'(g(x)) by g'(x) using the chain rule. f'(x) = sec^2(g(x)) * g'(x)

Substitute the values we found for g(x) and g'(x): f'(x) = sec^2(x + pi/4) * 1

Therefore, the derivative of f(x) = tg(x+pi/4) is f'(x) = sec^2(x + pi/4).

2) f(x) = sin(2x) + tg(x)

To find the derivative of f(x) = sin(2x) + tg(x), we can use the sum rule and the derivative of sin(x) and tg(x).

Step 1: Find the derivative of sin(2x). The derivative of sin(2x) is 2cos(2x).

Step 2: Find the derivative of tg(x). The derivative of tg(x) is sec^2(x).

Step 3: Combine the derivatives using the sum rule. f'(x) = 2cos(2x) + sec^2(x)

Therefore, the derivative of f(x) = sin(2x) + tg(x) is f'(x) = 2cos(2x) + sec^2(x).

3) f(x) = 4 - 1/4tg(x)

To find the derivative of f(x) = 4 - (1/4)tg(x), we can use the constant rule and the derivative of tg(x).

Step 1: Find the derivative of tg(x). The derivative of tg(x) is sec^2(x).

Step 2: Apply the constant rule. The derivative of a constant times a function is equal to the constant times the derivative of the function.

f'(x) = 0 - (1/4) * sec^2(x)

Therefore, the derivative of f(x) = 4 - (1/4)tg(x) is f'(x) = -(1/4)sec^2(x).

4) f(x) = x^2ctgx

To find the derivative of f(x) = x^2ctgx, we can use the product rule and the derivative of ctg(x) (which is -csc^2(x)).

Step 1: Find the derivative of x^2. The derivative of x^2 is 2x.

Step 2: Find the derivative of ctg(x). The derivative of ctg(x) is -csc^2(x).

Step 3: Apply the product rule. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by u'(x)v(x) + u(x)v'(x).

f'(x) = (2x * ctgx) + (x^2 * -csc^2(x))

Simplifying further: f'(x) = 2xctgx - x^2csc^2(x)

Therefore, the derivative of f(x) = x^2ctgx is f'(x) = 2xctgx - x^2csc^2(x).

I hope this helps! Let me know if you have any further questions.