Question #09e0a

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Question #09e0a

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Answer 1 · 2017-03-01T03:11:52

$\frac{d y}{d x} = 4 {tan}^{3} (x) {sec}^{2} (x) + 5 x^{4} {sec}^{2} (x^{5})$ $dy/dx = 4tan^3(x)sec^2(x) + 5x^4sec^2(x^5)$

Explanation:

Use the chain rule on each term.

First term: $y_{1} = {tan}^{4} (x)$ $y_1 = tan^4(x)$

Let $u = tan (x)$ $u = tan(x)$ , then $y = u^{4}, \frac{d y}{d u} = 4 u^{3}, and \frac{d u}{d x} = {sec}^{2} (x)$ $y = u^4, dy/(du)=4u^3, and (du)/dx = sec^2(x)$

$\frac{{d y}_{1}}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $dy_1/dx = dy/(du)(du)/dx$

$\frac{{d y}_{1}}{d x} = 4 u^{3} {sec}^{2} (x)$ $dy_1/dx = 4u^3sec^2(x)$

$\frac{{d y}_{1}}{d x} = 4 {tan}^{3} (x) {sec}^{2} (x)$ $dy_1/dx = 4tan^3(x)sec^2(x)$

Second term: $y_{2} = tan (x^{5})$ $y_2 = tan(x^5)$

Let $u = x^{2}$ $u = x^2$ , then $y = tan (u), \frac{d y}{d u} = {sec}^{2} (u), and \frac{d u}{d x} = 5 x^{4}$ $y = tan(u), dy/(du)=sec^2(u), and (du)/dx = 5x^4$

$\frac{{d y}_{2}}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $dy_2/dx = dy/(du)(du)/dx$

$\frac{{d y}_{2}}{d x} = {sec}^{2} (u) 5 x^{4}$ $dy_2/dx = sec^2(u)5x^4$

$\frac{{d y}_{2}}{d x} = 5 x^{4} {sec}^{2} (x^{5})$ $dy_2/dx = 5x^4sec^2(x^5)$

Put both terms back into the expression:

$\frac{d y}{d x} = \frac{{d y}_{1}}{d x} + \frac{{d y}_{2}}{d x}$ $dy/dx = dy_1/dx + dy_2/dx$

$\frac{d y}{d x} = 4 {tan}^{3} (x) {sec}^{2} (x) + 5 x^{4} {sec}^{2} (x^{5})$ $dy/dx = 4tan^3(x)sec^2(x) + 5x^4sec^2(x^5)$

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Question #09e0a

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