Question

Помогите срочно. Желательно с подробным решением.

Answer 1

1. $f' = \frac{1}{5} x^\frac{-4}{5} + 2*3x^2-4 = \frac{1}{5\sqrt[5]{x^4} } +6x^2-4\\f'(1) = \frac{1}{5} +6-4=\frac{11}{5}$

2. $f' = \frac{\frac{1}{2\sqrt{x} }(3x+1)-\sqrt{x} *3 }{(3x+1)^2} = \frac{\frac{3x+1-6x}{2\sqrt{x} } }{(3x+1)^2} = \frac{1-3x}{2\sqrt{x} (3x+1)^2} \\f'(2) = \frac{1-6}{2\sqrt{2} (6+1)^2} = \frac{-5}{98\sqrt{2} }$

3. $f' = 4*3x^2e^x-4x^3e^x = 12x^2e^x-4x^3e^x = 4x^2e^x(3-x)$

4. $f' = \frac{1}{3x^2-4x} (3*2x-4)=\frac{6x-4}{3x^2-4x}$

5. $f' = 4cos5x*5 = 20cos5x\\f'(\frac{\pi }{2} )= 20cos\frac{5\pi }{2} = 0$

6. $f' = \frac{1}{1+(\sqrt{6x-1})^2 } *\frac{1}{2\sqrt{6x-1} } *6=\frac{1}{2x\sqrt{6x-1} }$