Question

Differentiate a trig function that is greater than a power of 1, and involve either quotient, chain, or product rule.Differentiate a sine and cosine function that involves product and chain rule. Find the equation of the tangent line at x = a special triangle point (i.e. /4, /6, /3).Differentiate a function that involves both trig and exponential functions.[hint: add your own twist to this question for level 3/4]Differentiate an exponential function. [hint: add your own twist to this question for level 3/4]Differentiate a function where you have “y” and “x” on both sides of the equation and they cannot be simplified by collecting like terms or isolating y (i.e. y on one side and y^2 on the other). [hint: add your own twist to this question for level 3/4]

Answer 1

Given a trigonometric function that is greater than power of 1 as shown below:

`y=sin^2x\text{ ---- equation 1}`

To differentiate the function, we use the chain rule.

According to the chain rule,

`(dy)/(dx)=(dy)/(du)*(du)/(dx)`

From equation 1, let

`u=sin\text{ x --- equation 2}`

This implies that

`\begin{gathered} y=u^2 \\ \Rightarrow(dy)/(du)=2u \end{gathered}`

From equation 2,

`\begin{gathered} \begin{equation*} u=sin\text{ x} \end{equation*} \\ \Rightarrow(du)/(dx)=cos\text{ x} \end{gathered}`

`\begin{gathered} Recall\text{ from the chain rule:} \\ (dy)/(dx)=(dy)/(du)*(du)/(dx) \\ \Rightarrow2u*\text{cos x} \\ (dy)/(dx)=2ucos\text{ x} \\ but\text{ } \\ u=sin\text{ x} \\ \therefore(dy)/(dx)=2(sin\text{ }x)(cos\text{ }x) \end{gathered}`