# Find the derivative with respect to the independent variable. f(x)=9\cos x-2\sin

Find the derivative with respect to the independent variable.
$f\left(x\right)=9\mathrm{cos}x-2\mathrm{sin}x$
f'(x)=
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Step 1
We have to find derivatives of function:
$f\left(x\right)=9\mathrm{cos}x-2\mathrm{sin}x$
We know the derivatives of trigonometric ratios:
$\frac{d\mathrm{cos}x}{dx}=-\mathrm{sin}x$
$\frac{d\mathrm{sin}x}{dx}=\mathrm{cos}x$
Step 2
So differentiating the function $f\left(x\right)=9\mathrm{cos}x-2\mathrm{sin}x$ with respect to 'x', we get
$f\left(x\right)=9\mathrm{cos}x-2\mathrm{sin}x$
${f}^{\prime }\left(x\right)=9\frac{d\mathrm{cos}x}{dx}-2\frac{d\mathrm{sin}x}{dx}$
$=9\left(-\mathrm{sin}x\right)-2\left(\mathrm{cos}x\right)$
$=-9\mathrm{sin}-2\mathrm{cos}x$
Hence, value of .