Physics, asked by meghakatiyar1, 1 year ago

y(x) = (sin x + cosx)(cosx + sinx)

do differentiation of this function . ​

Answers

Answered by radhikaraodelhi06
2

Answer:

Explanation:

y

=

2

(

sin

x

cos

x

)

2

Explanation:

Start by taking a look at your function

y

=

sin

x

+

cos

x

sin

x

cos

x

Notice that this function is actually the quotient of two other functions, let's call them  

f

(

x

)

and  

g

(

x

)

{

f

(

x

)

=

sin

x

+

cos

x

g

(

x

)

=

sin

x

cos

x

This means that you can differentiate this function by using the quotient rule, which allows you to find the derivative of a function that's the quotient of two other functions by using the formula

d

d

x

(

y

)

=

f

(

x

)

g

(

x

)

f

(

x

)

g

(

x

)

[

g

(

x

)

]

2

, where  

g

(

x

)

0

You also need to remember that

d

d

x

(

sin

x

)

=

cos

x

and that

d

d

x

(

cos

x

)

=

sin

x

So, calculate the derivative of  

f

(

x

)

d

d

x

(

f

(

x

)

)

=

d

d

x

(

sin

x

+

cos

x

)

d

d

x

(

f

(

x

)

)

=

d

d

x

(

sin

x

)

+

d

d

x

(

cos

x

)

d

d

x

(

f

(

x

)

)

=

cos

x

sin

x

and that of  

g

(

x

)

d

d

x

(

g

(

x

)

)

=

d

d

x

(

sin

x

cos

x

d

d

x

(

g

(

x

)

)

=

d

d

x

(

sin

x

)

d

d

x

(

cos

x

)

d

d

x

(

g

(

x

)

)

=

cos

x

(

sin

x

)

=

cos

x

+

sin

x

The derivative of  

y

will thus look like this

d

d

x

(

y

)

=

(

cos

x

sin

x

)

(

sin

x

cos

x

)

(

sin

x

+

cos

x

)

(

cos

x

+

sin

x

)

(

sin

x

cos

x

)

2

I'll break this into two fractions to make the calculations easier to read

d

d

x

(

y

)

=

y

=

f

(

x

)

g

(

x

)

g

(

x

)

2

  

A

 

f

(

x

)

g

(

x

)

g

(

x

)

2

  

B

 

=

A

B

The first fraction will be equal to

A

=

(

cos

x

sin

x

)

(

sin

x

cos

x

)

(

sin

x

cos

x

)

2

A

=

cos

x

sin

x

cos

2

x

sin

2

x

+

sin

x

cos

x

(

sin

x

cos

x

)

2

A

=

2

sin

x

cos

x

(

sin

2

x

+

cos

2

x

)

(

sin

x

cos

x

)

2

You can further simplify this by using the fact that

sin

2

x

+

cos

2

x

=

1

A

=

2

sin

x

cos

x

1

(

sin

x

cos

x

)

2

The second fraction will be equal to

B

=

(

sin

x

+

cos

x

)

(

cos

x

+

sin

x

)

(

sin

x

cos

x

)

2

B

=

(

sin

x

+

cos

x

)

2

(

sin

x

cos

x

)

2

You can simplify this by using

(

a

+

b

)

2

=

a

2

+

2

a

b

+

b

2

to get

B

=

sin

2

x

+

2

sin

x

cos

x

+

cos

2

x

(

sin

x

cos

x

)

2

B

=

2

sin

x

cos

x

+

1

(

sin

x

cos

x

)

2

Put the two fractions back together to get

y

=

2

sin

x

cos

x

1

(

2

sin

x

cos

x

+

1

)

(

sin

x

cos

x

)

2

y

=

2

sin

x

cos

x

1

2

sin

x

cos

x

1

(

sin

x

cos

x

)

2

Finally, you get

y

=

2

(

sin

x

cos

x

)

2

Answered by loluzaka
0

Answer:

Answer:

n geometry, a triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides. A right triangular prism has rectangular sides, otherwise it is oblique. A uniform triangular prism is a right triangular prism with equilateral bases, and square sides.

Equivalently, it is a polyhedron of which two faces are parallel, while the surface normals of the other three are in the same plane (which is not necessarily parallel to the base planes). These three faces are parallelograms. All cross-sections parallel to the base faces are the same triangle.

Contents

1 As a semiregular (or uniform) polyhedron

2 Volume

3 Truncated triangular prism

4 Facetings

5 Related polyhedra and tilings

5.1 Symmetry mutations

5.2 Compounds

5.3 Honeycombs

5.4 Related polytopes

5.5 Four dimensional space

6 See also

7 References

As a semiregular (or uniform) polyhedron

A right triangular prism is semiregular or, more generally, a uniform polyhedron if the base faces are equilateral triangles, and the other three faces are squares. It can be seen as a truncated trigonal hosohedron, represented by Schläfli symbol t{2,3}. Alternately it can be seen as the Cartesian product of a triangle and a line segment, and represented by the product {3}x{}. The dual of a triangular hi prism is a triangular bipyramid.

The symmetry group of a right 3-sided prism with triangular base is D3h of order 12. The rotation group is D3 of order 6. The symmetry group does not contain inversion.

Volume

The volume of any prism is the product of the area of the base and the distance between the two bases. In this case the base is a triangle so we simply need to compute the area of the triangle and multiply this by the length of the prism:

{\displaystyle V={\frac {1}{2}}bhl,} {\displaystyle V={\frac {1}{2}}bhl,}

where b is the length of one side of the triangle, h is the length of an altitude drawn to that side, and l is the distance between the triangular faces.

Truncated triangular prism

A truncated right triangular prism has one triangular face truncated (planed) at an oblique angle.[1]

TruncatedTriangularPrism.png

Facetings

There are two full D2h symmetry facetings of a triangular prism, both with 6 isosceles triangle faces, one keeping the original top and bottom triangles, and one the original squares. Two lower C3v symmetry faceting have one base triangle, 3 lateral crossed square faces, and 3 isosceles triangle lateral faces.

Convex Facetings

D3h symmetry C3v symmetry

Triangular prism.png FacetedTriangularPrism2.png FacetedTriangularPrism.png FacetedTriangularPrism3.png FacetedTriangularPrism4.png

2 {3}

3 {4} 3 {4}

6 ( ) v { } 2 {3}

6 ( ) v { } 1 {3}

3 t'{2}

6 ( ) v { } 1 {3}

3 t'{2}

3 ( ) v { }

Related polyhedra and tilings

A regular tetrahedron or tetragonal disphenoid can be dissected into two halves with a central square. Each half is a topological triangular prism.

Family of uniform prisms [ vte ]

Family of convex cupolae [ vte ]

n 2 3 4 5 6

Name {2} || t{2} {3} || t{3} {4} || t{4} {5} || t{5} {6} || t{6}

Cupola Triangular prism wedge.png

Digonal cupola Triangular cupola.png

Triangular cupola Square cupola.png

Square cupola Pentagonal cupola.png

Pentagonal cupola Hexagonal cupola flat.png

Hexagonal cupola

(Flat)

Related

uniform

polyhedra Triangular prism

CDel node 1.pngCDel 2.pngCDel node.pngCDel 3.pngCDel node 1.png Cubocta-

hedron

CDel node 1.pngCDel 3.pngCDel node.pngCDel 3.pngCDel node 1.png Rhombi-

cubocta-

hedron

CDel node 1.pngCDel 4.pngCDel node.pngCDel 3.pngCDel node 1.png Rhomb-

icosidodeca-

hedron

CDel node 1.pngCDel 5.pngCDel node.pngCDel 3.pngCDel node 1.png Rhombi-

trihexagonal

tiling

CDel node 1.pngCDel 6.pngCDel node.pngCDel 3.pngCDel node 1.png

Symmetry mutations

This polyhedron is topologically related as a part of sequence of uniform truncated polyhedra with vertex configurations (3.2n.2n), and [n,3] Coxeter group symmetry.

Explanation:

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