Question

f(x)=\frac{\sqrt{2x^{2}}}{\cos(x)}
which sequence of rules can be used in order to differentiate f
in its current form?
choose 1 answer:
a product rule, then quotient rule
b chain rule, then chain rule again
c quotient rule, then quotient rule again
d quotient rule, then chain rule

Explanation:

Step1: Analyze the function form

The function \( f(x)=\frac{\sqrt{2x^{2}}}{\cos(x)} \) is a quotient of two functions \( u(x)=\sqrt{2x^{2}}=(2x^{2})^{\frac{1}{2}} \) and \( v(x)=\cos(x) \). So, we first use the quotient rule. The quotient rule states that if \( y = \frac{u}{v} \), then \( y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}} \).

Step2: Differentiate \( u(x)=(2x^{2})^{\frac{1}{2}} \)

For \( u(x)=(2x^{2})^{\frac{1}{2}} \), we use the chain rule. Let \( t = 2x^{2} \), then \( u(t)=t^{\frac{1}{2}} \). The chain rule states that if \( y = f(g(x)) \), then \( y^\prime=f^\prime(g(x))\cdot g^\prime(x) \). \( u^\prime(t)=\frac{1}{2}t^{-\frac{1}{2}} \) and \( t^\prime = 4x \), so \( u^\prime(x)=\frac{1}{2}(2x^{2})^{-\frac{1}{2}}\cdot4x\). \( v^\prime(x)=-\sin(x) \) (derivative of \( \cos(x) \) is \( -\sin(x) \)).

Answer:

D. Quotient rule, then chain rule