# Question #af0e0

Aug 4, 2017

$\cos \left(x\right) + {\sec}^{2} \left(x\right)$

#### Explanation:

Since the derivative of a sum is the sum of the derivatives

$\left(\sin \left(x\right) + \tan \left(x\right)\right) ' = \left(\sin \left(x\right)\right) ' + \left(\tan \left(x\right)\right) '$

Also, recall that $\left(\sin \left(x\right)\right) ' = \cos \left(x\right)$

And $\left(\cos \left(x\right)\right) ' = - \sin \left(x\right)$

Then

$\left(\sin \left(x\right) + \tan \left(x\right)\right) ' = \cos \left(x\right) + \left(\tan \left(x\right)\right) '$

Since $\tan \left(x\right) = \frac{\sin \left(x\right)}{\cos \left(x\right)}$

Then $\left(\tan \left(x\right)\right) ' = \frac{{\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)}{{\cos}^{2} \left(x\right)}$ by the quotient rule

And since ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

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

Then

$\left(\sin \left(x\right) + \tan \left(x\right)\right) ' = \cos \left(x\right) + {\sec}^{2} \left(x\right)$