Question

1. $f(x)=\tan\sqrt{x}$

j. $\sec^{2}(\sec^{2}\sqrt{x})(\frac{1}{2\sqrt{x}})$
c. $(\sec\sqrt{x})(\frac{1}{2\sqrt{x}})$
e. $(\sec^{2}\sqrt{x})(\frac{1}{2\sqrt{x}})$
u. $(\sec\sqrt{x})(\frac{1}{\sqrt{x}})$

Explanation:

Step1: Apply the chain rule

The chain rule states that if \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). Let \(u = \sqrt{x}=x^{\frac{1}{2}}\), so \(f(x)=\tan(u)\). The derivative of \(\tan(u)\) with respect to \(u\) is \(\sec^{2}(u)\), and the derivative of \(u = x^{\frac{1}{2}}\) with respect to \(x\) is \(u^\prime=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\).

Step2: Substitute back

By the chain rule \(f^\prime(x)=\sec^{2}(\sqrt{x})\cdot\frac{1}{2\sqrt{x}}\)

Answer:

E. \((\sec^{2}\sqrt{x})(\frac{1}{2\sqrt{x}})\)