Question

F(x)=3/x^9-x^8/4+4√x f(x)=sin x*(3-x^2) f(x)= x+6/cos x f(x)=(2x^8-3)^15 f(x)=sin 3x

Answer 1

1)f`(x)=3*(-9)*x^(-10)-8x^7/4+4*1/2*x^(-1/2)=-27/x^10-2x^7+2/Vx

V-знак корня

2)f`(x)=cosx(3-x^2)+sinx(-2x)=cosx(3-x^2)-2x*sinx

3)f`(x)=(cosx-(x+6)(-sinx))/cos^2x=(cosx+(x+6)sinx)/cos^2x

4)f`(x)=15(2x^8-3)^14*16x^7=240x^7(2x^8-3)^14

5)f`(x)=cos3x*3=3cos3x

Answer 2

To differentiate the given functions, we'll use the standard rules of calculus. Let's find the derivatives of each function:

1. F(x) = 3/x^9 - x^8/4 + 4√x

Using the power rule, we can differentiate each term of the function separately: d/dx (3/x^9) = -27/x^10 d/dx (-x^8/4) = -2x^7 d/dx (4√x) = 2/x^(1/2)

Now, adding up the derivatives of all the terms: F'(x) = -27/x^10 - 2x^7 + 2/x^(1/2)

2. f(x) = sin(x) * (3 - x^2)

Using the product rule: d/dx [u(x) * v(x)] = u'(x) * v(x) + u(x) * v'(x)

Let u(x) = sin(x) and v(x) = (3 - x^2). Then, u'(x) = cos(x) and v'(x) = -2x.

Now, applying the product rule: f'(x) = cos(x) * (3 - x^2) + sin(x) * (-2x) f'(x) = 3cos(x) - x^2cos(x) - 2xsin(x)

3. f(x) = (x + 6) / cos(x)

Using the quotient rule: d/dx [u(x) / v(x)] = (u'(x)v(x) - v'(x)u(x)) / (v(x))^2

Let u(x) = (x + 6) and v(x) = cos(x). Then, u'(x) = 1 and v'(x) = -sin(x).

Now, applying the quotient rule: f'(x) = (1 * cos(x) - (-sin(x)) * (x + 6)) / cos^2(x) f'(x) = (cos(x) + (x + 6)sin(x)) / cos^2(x)

4. f(x) = (2x^8 - 3)^15

Using the chain rule: d/dx [u(v(x))] = u'(v(x)) * v'(x)

Let u(x) = x^15 and v(x) = 2x^8 - 3. Then, u'(x) = 15x^14 and v'(x) = 16x^7.

Now, applying the chain rule: f'(x) = 15(2x^8 - 3)^14 * 16x^7 f'(x) = 240x^7(2x^8 - 3)^14

5. f(x) = sin(3x)

Using the chain rule: d/dx [u(v(x))] = u'(v(x)) * v'(x)

Let u(x) = sin(x) and v(x) = 3x. Then, u'(x) = cos(x) and v'(x) = 3.

Now, applying the chain rule: f'(x) = cos(3x) * 3 f'(x) = 3cos(3x)

These are the derivatives of the given functions with respect to x.