2 ответа:
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<span>1) f(x)=√3x+1, x0=5;
</span>
<span>f'(5) = 3/(2sqrt(16)) = 3/8
2) f(x) = sin^5x, x0=π/3
f'(x) = 5cos(5x)
f'(</span>π/3) = 5cos(5π/3) = 5cos(2π-π/3) = 5cos(π/3) = 5/2
</span>
<span>f'(5) = 3/(2sqrt(16)) = 3/8
2) f(x) = sin^5x, x0=π/3
f'(x) = 5cos(5x)
f'(</span>π/3) = 5cos(5π/3) = 5cos(2π-π/3) = 5cos(π/3) = 5/2