Please find the derivative dy/dx of the following function. Is this right? g) tan(x - y) =1+x²d1/² (tan ex-x) = // (1+x)dxdxCímoSec² (x-y)- $ x (x²-y) = (1 + x) · X · Y. √x (1+x)dxSec² (x-y). (1-dy) =(1+x).(1+x²2/x-y. (0+2x)dySera (x-y)- Seca (x-y).(1+x² + sec ³ (x-1))seco(x-1)+ axy(1+x)(1+x)2• d = 1 1dxTh) Secx=xtany+ sec (X-Y)Final AnswerSecalx-x + xy(1+x²)+ Sei CX-Y)+ x³· dy + sec³(x-1). dx = sec ² (x-x)+ axyCittaja1+x²Hdy.XYJx (1 + x)²sec (x-1) + 2xy(1+x)Final Answer

Given query is to find the derivative.

dv g) tan(x- y) = y 1+x² d 1/₂ (tan ex-x) = $x (1+x) dx dx Sec² (x-y). (1-dy) = dx Sec ³ (x-y) - dy + d sec² (x-y) · —— (x²-y) = (1 + x) · ** - Y- £x (1-x) dy dx (1+x²) Sec²³(x-1). dx = dy dx (1+x²³² = (1+x). Jy - y. (0 +ax) (1+x). (1+x) (1+x² + sec ³ (x-1)). 2 sec³(x-4) + 2xy sec ³ (x-Y). dy Jx = dr 1 Final Answe Seca Ix-X + Se. dy (1 + x²³) Jx = seca (x-y)+ axy Cit dy = sel³ (x-1) + axy. Se (1+x) - CI

dv g) tan(x- y) = y 1+x² d 1/₂ (tan ex-x) = $x (1+x) dx dx Sec² (x-y). (1-dy) = dx Sec ³ (x-y) - dy + d sec² (x-y) · —— (x²-y) = (1 + x) · ** - Y- £x (1-x) dy dx (1+x²) Sec²³(x-1). dx = dy dx (1+x²³² = (1+x). Jy - y. (0 +ax) (1+x). (1+x) (1+x² + sec ³ (x-1)). 2 sec³(x-4) + 2xy sec ³ (x-Y). dy Jx = dr 1 Final Answe Seca Ix-X + Se. dy (1 + x²³) Jx = seca (x-y)+ axy Cit dy = sel³ (x-1) + axy. Se (1+x) - CI

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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