Question

solv.... the math.....plz​

Answer 1

⠀⠀|| ☆.Question.☆ ||

Find the derivative of (x²+1).Sin x .

⠀ ⠀|| ☆.Solution.☆ ||

Find :-

• Derivative of (x²+1).Sin x

⠀|| ☆.Explanation.☆ ||

We know,

★ d sin x/dx = cos x.

★ d/dx (x^n) = n.d/dx[x^(n-1)]

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So,

➥d/dx [(x²+1).sin x]

By part of derivative

If, y = pq

So, derivative will be

➥ dy/dx = d/dx(pq)

➥ dy/dx = p.dq/dx + q.dp/dx

So,

➥(x²+1) d/dx(sin x) + sin x d/dx(x²+1)

➥(x²+1).cos x + sin x .(2x+0)

➥cos x(x²+1) + 2x. sin x

Or,

➥ x². cos x + 2x . sin x + cos x [Ans]

Some important derivative

★d/dx(sin x) = cos x

★ d/dx(cos x) = -sin x

★ d/dx(tan x) = sec² x

★d/dx(sec x) = sec x.tan x

★ d/dx(cot x) = - cosec² x

★ d/dx(cosec x) = -cosec x.cot x

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Answer 2

AnswEr :

Given that,

$\sf \: y = ( {x}^{2} + 1)sin \: x$

We have to find the derivative of above expression

Formulas Used :

The above expression is in the form uv

Derivative of the above kind of expressions would be of the form :

• $\large \sf \: y' = u'v + v'u$

This is known as the Product Rule of Differentiation

• $\large \sf x^n = nx^{n - 1}$

This is known as the Power Rule of Differentiation

$\rule{300}{2}$

Differentiating the above expression w.r.t x ,

$\longrightarrow \: \sf \: y' = \dfrac{d( {x}^{2} + 1)}{dx} \times sin \: x + \dfrac{d(sin \: x)}{dx} \: ( {x}^{2} + 1) \\ \\ \longrightarrow \: \sf \: y' = \: 2x \: sin \: x + cos \: x( {x}^{2} + 1)$

• Derivative of x² + 1 is 2x

• Derivative of sin x is cos x

The derivative of y would be 2x sin x + (x² + 1)cos x

$\rule{300}{2}$

$\rule{300}{2}$