Question

Помогите с производными

Answer 1

a)

$y = (5x+4)^7\\f'(x) = ((5x+4)^7)'*(5x+4)' = 7(5x+4)^6*5 = 35(5x+4)^6$

б)

$y = \sqrt{9-2x}.\\f'(x) = (\sqrt{9-2x})'*(9-2x)' = \frac{1}{2*\sqrt{9-2x}}*-2 = -\frac{1}{\sqrt{9-2x}}$

в)

$y = ln(5-2x)\\f'(x) = (ln(5-2x))'*(5-2x)' = \frac{1}{5-2x}*-2 = - \frac{2}{5-2x}$

г)

$y = tg(lnx)\\f'(x) = (tg(lnx))'*(lnx)' = \frac{1}{cos^2(lnx)}*\frac{1}{x} = \frac{1}{xcos^2(lnx))}$

д)

$y = cos(4x+\frac{\pi }{3})\\f'(x) = (cos(4x+\frac{\pi}{3}))'*(4x+\frac{\pi}{3})' = -sin(4x+\frac{\pi}{3})*4 = -4sin(4x+\frac{\pi}{3})$

e)

$y = arcsin(4x)\\f'x = (arcsin(4x))'*(4x)' = \frac{1}{\sqrt{1-16x^2}}*4 = \frac{4}{\sqrt{1-16x^2}}$

ж)

$y=e^{cosx}\\f'(x) = (e^{cosx})'*cosx' = e^{cosx}*-sinx = -e^{cosx}sinx$

з)

$y = ln(x^2-5x)\\f'(x) = (ln(x^2-5x))'*(x^2-5x)' = \frac{1}{x^2-5x}*(2x-5) = \frac{2x-5}{x^2-5x}$

и)

$y = (3x+8)^6\\f'(x) = ((3x+8)^6)'*(3x+8)' = 6(3x+8)^5*3 = 18(3x+8)^5$

к)

$y = \frac{1}{(6x-1)^3} = (6x-1)^{-3}\\f'(x) = ((6x-1)^{-3})'*(6x-1)' = -3*(6x-1)^{-4}*6 = -18*(6x-1)-4 = -\frac{18}{(6x-1)^4}$

л)

$y=arctg(\frac{\pi}{3}-x)\\f'(x) = (arctg(\frac{\pi}{3}-x))'*(\frac{\pi}{3}-x)' = \frac{1}{1+(\frac{\pi}{3}-x)^2}*-1 = -\frac{1}{(\frac{\pi}{3}-x)^2+1}$

м)

$y = 10^{ctgx}\\f'(x) = (10^{ctgx})'*ctgx' = 10^{ctgx}(-ctgx^2-1)ln10$