Question

if (f(x)=6sin x + 5cos x), then (f(x)=-6cos x - 5sin x) (f(1)=)

Explanation:

Step1: Recall derivative rules

We know that $\frac{d}{dx}(\sin x)=\cos x$ and $\frac{d}{dx}(\cos x)=-\sin x$. Also, by the sum - rule of derivatives $\frac{d}{dx}(u + v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)$ for functions $u$ and $v$. Given $f(x)=6\sin x + 5\cos x$, then $f'(x)=6\frac{d}{dx}(\sin x)+5\frac{d}{dx}(\cos x)$.

Step2: Calculate $f'(x)$

$f'(x)=6\cos x-5\sin x$.

Step3: Evaluate $f'(1)$

Substitute $x = 1$ (assuming $x$ is in radians) into $f'(x)$. So $f'(1)=6\cos(1)-5\sin(1)$. Using a calculator, $\cos(1)\approx0.5403$ and $\sin(1)\approx0.8415$. Then $f'(1)=6\times0.5403 - 5\times0.8415=3.2418-4.2075=- 0.9657$.

Answer:

$-0.9657$