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Chapter 3: Problem 66

Use the Chain Rule combined with other differentiation rules to find the derivative of the following functions. $$y=(z+4)^{3} \tan z$$

Short Answer

Expert verified
Question: Find the derivative of the function $$y = (z+4)^3 \tan z$$. Answer: The derivative of the function $$y = (z+4)^3 \tan z$$ is $$\frac{dy}{dz} = (z+4)^3 \cdot \sec^2 z + \tan z \cdot 3(z+4)^2$$.

Step by step solution

01

Identify u, v, du/dz, and dv/dz

First, identify the functions $$u$$ and $$v$$, and then differentiate them with respect to $$z$$ respectively. Let $$u = (z+4)^3$$ and $$v = \tan z$$. Now find the derivative of $$u$$ and $$v$$ with respect to $$z$$: $$\frac{du}{dz} = \frac{d(z+4)^3}{dz}, \quad \frac{dv}{dz} = \frac{d\tan z}{dz}$$
02

Differentiate u and v using the Chain Rule

Differentiate $$u$$ and $$v$$ using the Chain Rule. For $$\frac{du}{dz}$$: Let $$w = z+4$$, then $$u = w^3$$. $$\frac{du}{dw} = \frac{d(w^3)}{dw} = 3w^2$$ $$\frac{dw}{dz} = \frac{d(z+4)}{dz} = 1$$ Applying the Chain Rule: $$\frac{du}{dz} = \frac{du}{dw} \cdot \frac{dw}{dz} = 3w^2 \cdot 1=3(z+4)^2$$ For $$\frac{dv}{dz}$$: Since we know the derivative of $$\tan z$$ is $$\sec^2 z$$, we have $$\frac{dv}{dz} = \frac{d\tan z}{dz} = \sec^2 z$$
03

Apply the Product Rule

Now we have all the components needed to apply the Product Rule. $$\frac{dy}{dz} = u \cdot \frac{dv}{dz} + v \cdot \frac{du}{dz}$$ Substitute the values found in Steps 1 and 2: $$\frac{dy}{dz} = (z+4)^3 \cdot \sec^2 z + \tan z \cdot 3(z+4)^2$$ And this is the final answer for the derivative of the function $$y=(z+4)^{3} \tan z$$.

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Most popular questions from this chapter

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