How do you find the first and second derivative of sin^2(lnx)?

May 9, 2016

Use of chain rule twice and at the second derivative use of quotent rule.

First derivative

$2 \sin \left(\ln x\right) \cdot \cos \left(\ln x\right) \cdot \frac{1}{x}$

Second derivative

$\frac{2 \cos \left(2 \ln x\right) - \sin \left(2 \ln x\right)}{x} ^ 2$

Explanation:

First derivative

$\left({\sin}^{2} \left(\ln x\right)\right) '$

$2 \sin \left(\ln x\right) \cdot \left(\sin \left(\ln x\right)\right) '$

$2 \sin \left(\ln x\right) \cdot \cos \left(\ln x\right) \left(\ln x\right) '$

$2 \sin \left(\ln x\right) \cdot \cos \left(\ln x\right) \cdot \frac{1}{x}$

Although this is acceptable, to make the second derivative easier, one can use the trigonometric identity:

2sinθcosθ=sin(2θ)

Therefore:

$\left({\sin}^{2} \left(\ln x\right)\right) ' = \sin \frac{2 \ln x}{x}$

Second derivative

$\left(\sin \frac{2 \ln x}{x}\right) '$

$\frac{\sin \left(2 \ln x\right) ' x - \sin \left(2 \ln x\right) \left(x\right) '}{x} ^ 2$

$\frac{\cos \left(2 \ln x\right) \left(2 \ln x\right) ' x - \sin \left(2 \ln x\right) \cdot 1}{x} ^ 2$

$\frac{\cos \left(2 \ln x\right) \cdot 2 \cdot \frac{1}{x} \cdot x - \sin \left(2 \ln x\right)}{x} ^ 2$

$\frac{2 \cos \left(2 \ln x\right) - \sin \left(2 \ln x\right)}{x} ^ 2$