if $f(x)=7sin x + 11cos x$, then $f(x)=$ $f(3)=$

Explanation:

Step1: Recall derivative rules

The derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$. Using the sum - rule of derivatives $(u + v)'=u'+v'$, if $f(x)=7\sin x + 11\cos x$, then $f'(x)=\frac{d}{dx}(7\sin x)+\frac{d}{dx}(11\cos x)$. $f'(x)=7\cos x-11\sin x$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. The derivative of $\cos x$ is $-\sin x$ and the derivative of $\sin x$ is $\cos x$. So $f''(x)=\frac{d}{dx}(7\cos x)-\frac{d}{dx}(11\sin x)$. $f''(x)=- 7\sin x-11\cos x$.

Step3: Evaluate $f'(3)$

Substitute $x = 3$ into $f'(x)$. Since $f'(x)=7\cos x-11\sin x$, then $f'(3)=7\cos(3)-11\sin(3)$. Using a calculator (assuming $x$ is in radians), $\cos(3)\approx - 0.990$, $\sin(3)\approx0.141$. $f'(3)=7\times(-0.990)-11\times0.141=-6.93 - 1.551=-8.481$.

Answer:

$f''(x)=-7\sin x - 11\cos x$ $f'(3)\approx - 8.481$