Question
Asked Sep 26, 2019

Using the chain rule find dy/dx for the function:

y=sin(cos(tan(sec(csc(cot(e^((2x^2)(2x)(2)))))))))

check_circleExpert Solution
Step 1

Given:

y sin(cos(tan(sec(csc(cot(2(2x*>»(2><2)))
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y sin(cos(tan(sec(csc(cot(2(2x*>»(2><2)))

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Step 2

Formula used:

(sinx) cosx
dx
d
(cos.x) sinx
dx
d
(tan x) sec2 x
dx
(secx) secx tan x
dx
d
(cot x)csc2x
dx
-(cscx)-cscxcotx
dx
11
d
e =e*
dx
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(sinx) cosx dx d (cos.x) sinx dx d (tan x) sec2 x dx (secx) secx tan x dx d (cot x)csc2x dx -(cscx)-cscxcotx dx 11 d e =e* dx

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Step 3

So, using chain rule the derivative of ‘y&rsquo...

>(2))
y =sin(cos(tan(sec(csc (cot(e2x*>2xX2}
Sx3
'»]x
sincos(tan(sec(csc(cot(e
dx
d
cos(tan(sec(csc(cot(e? x
dx
dy
d
tan(sec(csc(cot(e
dx
x
dx
d
d
cot(e
dx
$x
8x3
dx
dx
d
dx
|
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>(2)) y =sin(cos(tan(sec(csc (cot(e2x*>2xX2} Sx3 '»]x sincos(tan(sec(csc(cot(e dx d cos(tan(sec(csc(cot(e? x dx dy d tan(sec(csc(cot(e dx x dx d d cot(e dx $x 8x3 dx dx d dx |

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