Пример: <span>f(x)=7+5⋅cos<span>(3⋅x) </span></span><span><span>f′</span>(x)=<span><span>(7+5⋅cos<span>(3⋅x)</span>)</span>′</span>=</span><span>=<span><span>(5⋅cos<span>(3⋅x)</span>)</span>′</span>=</span><span>=5⋅<span><span>(cos<span>(3⋅x)</span>)</span>′</span>=</span><span>=5⋅<span>(−sin<span>(3⋅x)</span>)</span>⋅<span><span>(3⋅x)</span>′</span>=</span><span>=5⋅<span>(−sin<span>(3⋅x)</span>)</span>⋅3=</span><span>=15⋅<span>(−sin<span>(3⋅x)</span>)</span></span>Ответ:<span><span>f′</span>(x)=15⋅<span>(−sin<span>(3⋅x)</span>) Если скобки правильно поставила, то так. Если нет, то напишите правильно пример.</span></span>