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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]

Differentiate A Trig Function That Is Greater Than A Power Of 1 And Involve Either Quotient Chain Or Product RuleDifferentiate A Sine And Cosine Function That I class=
Differentiate A Trig Function That Is Greater Than A Power Of 1 And Involve Either Quotient Chain Or Product RuleDifferentiate A Sine And Cosine Function That I class=

Sagot :

Solution:

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

[tex]y=sin^2x\text{ ---- equation 1}[/tex]

To differentiate the function, we use the chain rule.

According to the chain rule,

[tex]\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}[/tex]

From equation 1, let

[tex]u=sin\text{ x --- equation 2}[/tex]

This implies that

[tex]\begin{gathered} y=u^2 \\ \Rightarrow\frac{dy}{du}=2u \end{gathered}[/tex]

From equation 2,

[tex]\begin{gathered} \begin{equation*} u=sin\text{ x} \end{equation*} \\ \Rightarrow\frac{du}{dx}=cos\text{ x} \end{gathered}[/tex]

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

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