# D (log sin cos tan e^cosx)/dx ?

Jul 9, 2018

$= \cot \left(\cos \left(\tan {e}^{\cos} x\right)\right) \cdot \sin \left(\tan \left({e}^{\cos} x\right)\right) \cdot {\sec}^{2} \left({e}^{\cos} x\right) \cdot {e}^{\cos} x \cdot \sin x$

#### Explanation:

The chain rule for a single composition is:

• $\left({f}_{1} \circ {f}_{2}\right) ' = {f}_{1} ' {f}_{2} '$

The Lagrange/ prime notation of the RHS signifies differentiation with respect to the function's own argument.

Extending for multiple composition:

• $\left({f}_{1} \circ {f}_{2} \circ {f}_{3.} \ldots\right) ' = {f}_{1} ' {f}_{2} ' {f}_{3} ' \ldots$

For:  log (sin (cos (tan (e^cosx)))))

• Let: $\left\{\begin{matrix}{a}_{1} \left(x\right) = \sin \left(\cos \left(\tan \left({e}^{\cos} x\right)\right)\right) \\ {a}_{2} \left(x\right) = \cos \left(\tan \left({e}^{\cos} x\right)\right) \\ {a}_{3} \left(x\right) = \tan \left({e}^{\cos} x\right) \\ {a}_{4} \left(x\right) = {e}^{\cos} x \\ {a}_{5} \left(x\right) = \cos x\end{matrix}\right.$

$\implies \frac{d}{\mathrm{dx}} \left(\log \left(\sin \left(\cos \left(\tan \left({e}^{\cos} x\right)\right)\right)\right)\right) q \quad \square$

$= {\underbrace{\frac{1}{a} _ 1}}_{= {\left(\log {a}_{1}\right)}^{'}} \cdot \cos {a}_{2} \cdot \left(- \sin {a}_{3}\right) \cdot {\sec}^{2} {a}_{4} \cdot {a}_{4} \cdot {\underbrace{- \sin x}}_{= {a}_{5} '}$

$= \frac{1}{\sin \left(\cos \left(\tan \left({e}^{\cos} x\right)\right)\right)} \cdot \cos \left(\cos \left(\tan \left({e}^{\cos} x\right)\right)\right) \cdot \sin \left(\tan \left({e}^{\cos} x\right)\right) \cdot {\sec}^{2} \left({e}^{\cos} x\right) \cdot {e}^{\cos} x \cdot \sin x$

$= \cot \left(\cos \left(\tan \left({e}^{\cos} x\right)\right)\right) \cdot \sin \left(\tan \left(\left({e}^{\cos} x\right)\right)\right) \cdot {\sec}^{2} \left({e}^{\cos} x\right) \cdot {e}^{\cos} x \cdot \sin x$

In Liebnitz notation:

• $\square = \frac{d \left(\log {a}_{1}\right)}{d {a}_{1}} \frac{d {a}_{1}}{d {a}_{2}} \frac{d {a}_{2}}{d {a}_{3}} \frac{d {a}_{3}}{d {a}_{4}} \frac{d {a}_{4}}{d {a}_{5}} \frac{d {a}_{5}}{\mathrm{dx}} \boldsymbol{= \frac{d \left(\log {a}_{1}\right)}{d x}}$