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Chapter 3: Problem 36

Find the derivative of the following functions. $$y=\frac{\tan w}{1+\tan w}$$

Short Answer

Expert verified
Question: Find the derivative of the function $$y=\frac{\tan w}{1+\tan w}$$ with respect to w. Answer: The derivative of the function $$y=\frac{\tan w}{1+\tan w}$$ with respect to $$w$$ is $$y'(w)=\frac{\sec^2 w}{(1+\tan w)^2}$$.

Step by step solution

01

Quotient Rule

When finding the derivative of a quotient of two functions, we apply the quotient rule, which states that if $$y=\frac{u(w)}{v(w)}$$, then: $$y'(w) = \frac{u'(w) \cdot v(w) - u(w) \cdot v'(w)}{[v(w)]^2}$$. In our case we have: $$u(w)=\tan w$$ and $$v(w)=1+\tan w$$.
02

Derivatives of u and v with respect to w

We need to compute $$u'(w)$$ and $$v'(w)$$. The derivative of $$\tan w$$ with respect to w is $$\sec^2 w$$. So, $$u'(w)=\sec^2 w$$. The derivative of $$1 + \tan w$$ with respect to w is again $$\sec^2 w$$. So, $$v'(w)=\sec^2 w$$.
03

Substitute u', v', u and v into Quotient Rule

Now, we substitute $$u'(w)$$, $$v'(w)$$, $$u(w)$$, and $$v(w)$$ into the quotient rule formula: $$y鈥�(w) = \frac{u'(w) \cdot v(w) - u(w) \cdot v'(w)}{[v(w)]^2} = \frac{(\sec^2 w)(1+\tan w) - (\tan w)(\sec^2 w)}{(1+\tan w)^2}$$
04

Simplify the Derivative

Simplify the expression: $$y'(w) = \frac{\sec^2 w + \sec^2 w \tan w - \tan w \sec^2 w}{(1+\tan w)^2} = \frac{\sec^2 w}{(1+\tan w)^2}$$
05

Final Answer: The derivative of the function

The derivative of the function $$y=\frac{\tan w}{1+\tan w}$$ with respect to $$w$$ is: $$y'(w)=\frac{\sec^2 w}{(1+\tan w)^2}$$

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Most popular questions from this chapter

Given the function \(f,\) find the slope of the line tangent to the graph of \(f^{-1}\) at the specified point on the graph of \(f^{-1}\) . $$f(x)=-x^{2}+8 ;(7,1)$$

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