How do you differentiate f(x)=sin2x * cot(5-x) using the product rule?

Mar 29, 2018

$2 \cos \left(2 x\right) \cot \left(5 - 2 x\right) + \sin \left(2 x\right) \cot \left(5 - x\right) \csc \left(5 - x\right)$

Explanation:

The rule for differentiating a product of two functions is

$\frac{d}{\mathrm{dx}} \left(u v\right) = \frac{\mathrm{du}}{\mathrm{dx}} v + u \frac{\mathrm{dv}}{\mathrm{dx}}$

Thus

$f \left(x\right) = \sin 2 x \cdot \cot \left(5 - x\right) q \quad \implies q \quad$

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} \left(\sin 2 x\right) \cdot \cot \left(5 - x\right) + \sin 2 x \cdot \frac{d}{\mathrm{dx}} \left(\cot \left(5 - x\right)\right)$
$q \quad = \left(2 \cos \left(2 x\right)\right) \cdot \cot \left(5 - 2 x\right) + \sin \left(2 x\right) \cdot \left(- \cot \left(5 - x\right) \csc \left(5 - x\right) \times \left(- 1\right)\right)$
$q \quad = 2 \cos \left(2 x\right) \cot \left(5 - 2 x\right) + \sin \left(2 x\right) \cot \left(5 - x\right) \csc \left(5 - x\right)$

where we have used the standard rules

$\frac{d}{\mathrm{dx}} \sin x = \cos x , q \quad \frac{d}{\mathrm{dx}} \cot x = - \cot x \csc x$

Mar 29, 2018

The derivative of $\sin 2 x \cot \left(5 - x\right)$ with respect to $x$ is
$2 \cos 2 x \cot \left(5 - x\right) + \sin 2 x {\csc}^{2} \left(5 - x\right)$.

Explanation:

The product rule states

$\frac{d \left[f \left(x\right) \cdot g \left(x\right)\right]}{\mathrm{dx}} = \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} g \left(x\right) + f \left(x\right) \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}$

Here, $f \left(x\right) = \sin 2 x$, and $g \left(x\right) = \cot \left(5 - x\right)$

Note that $f \left(x\right)$ and $g \left(x\right)$ are both COMPOSITE functions, so to solve this problem we will also need to use the chain rule which states that

$\frac{\mathrm{df} \left(g \left(x\right)\right)}{\mathrm{dx}} = {f}^{'} \left(g \left(x\right)\right) \cdot {g}^{'} \left(x\right)$.

Now that we know the rules, let's compute the derivative

$\frac{\mathrm{ds} \in 2 x \cdot \cot \left(5 - x\right)}{\mathrm{dx}} = \frac{\mathrm{ds} \in 2 x}{\mathrm{dx}} \cot \left(5 - x\right) + \sin 2 x \frac{\mathrm{dc} o t \left(5 - x\right)}{\mathrm{dx}}$

$= 2 \cos 2 x \cot \left(5 - x\right) + \sin 2 x \left[- {\csc}^{2} \left(5 - x\right) \cdot \left(- 1\right)\right]$

$= 2 \cos 2 x \cot \left(5 - x\right) + \sin 2 x {\csc}^{2} \left(5 - x\right)$