Question

obtain derivatives of the following functions 1) xsin x 2) x4 + cos x 3) X/sinx

Answer 1

Explanation:

1) sin x + x(cos x)

2) 4x^3

3) ( sinx - x.cosx ) / (sinx)^2

hope it helps uh.....

Answer 2

Answer:

1.$\frac{d(xsinx}{dx}=sinx +xcosx$

2.$\frac{d(x^4+cos x^3)}{dx}=4x^3-3x^2cosx$

3.$\frac{d(\frac{x}{sinx})}{dx}=\frac{sinx-xcosx}{sin^2x}$

Explanation:

We are given that

1.$xsinx$

We have to differentiate w.r.t x

We know that $\frac{d(u\cdot v)}{dx}=u'v+uv'$

$\frac{dx^n}{dx}=nx^{n-1},\frac{dsinx}{dx}=cosx,\frac{dcosx}{dx}=-sinx$,

Using the formula and differentiate w.r.t x

$\frac{d(xsinx}{dx}=1\times sinx+xcosx =sinx +xcos x$

$\frac{d(xsinx}{dx}=sinx +xcosx$

2.$\frac{d(x^4+cos x^3)}{dx}=4x^3+(-sinx^3)(3x^2)$

Using the rule PBA: power base angle in trigonometric function

$\frac{d(x^4+cos x^3)}{dx}=4x^3-3x^2cosx$

3.$\frac{x}{sinx}$

We have to differentiate w.r.t x

$\frac{d(\frac{u}{v})}{dx}=\frac{u'v-v'u}{v^2}$

Using this rule and differentiate w.r.t.x

Then, we get

$\frac{f(\frac{x}{sinx})}{dx}=\frac{1\cdot sinx-xcosx}{sin^2}$

$\frac{d(\frac{x}{sinx})}{dx}=\frac{sinx-xcosx}{sin^2x}$