What is the derivative of the following funct ion? (1) y=4sin^-1(3×) (2) y=ln(5×+sin^-1(×) (3) y=3cos(sin^-1(2×+3))

2 Answers
Mar 3, 2018

(1)(1) is 4/sqrt(1-x^2)41x2

(2)(2) is (5sqrt(1-x^2)+1)/((5x+sin^-1(x))sqrt(1-x^2))51x2+1(5x+sin1(x))1x2

(3)(3) is -(12x+18)/(isqrt(4x^2-12x-10))12x+18i4x212x10

Explanation:

(1)(1):

d/dx(4sin^-1(x))ddx(4sin1(x))

4*d/dxsin^-1(x)4ddxsin1(x)

4*1/sqrt(1-x^2)411x2

4/sqrt(1-x^2)41x2

(2)(2):

d/dx(ln(5x+sin^-1(x)))ddx(ln(5x+sin1(x)))

According to the chain rule, (df)/dx=(df)/(du)*(du)/(dx)dfdx=dfdududx, where uu is a function within ff.

Here, we have:

d/(du)ln(u)*d/dx(5x+sin^-1(x))dduln(u)ddx(5x+sin1(x))

1/u*(5+1/sqrt(1-x^2))1u(5+11x2)

But since u=5x+sin^-1(x)u=5x+sin1(x), we have:

1/(5x+sin^-1(x))(5+1/sqrt(1-x^2))15x+sin1(x)(5+11x2)

5/(5x+sin^-1(x))+1/(5x+sin^-1(x)sqrt(1-x^2))55x+sin1(x)+15x+sin1(x)1x2

(5sqrt(1-x^2)+1)/((5x+sin^-1(x))sqrt(1-x^2))51x2+1(5x+sin1(x))1x2

(3)(3):

d/dx(3cos(sin^-1(2x+3)))ddx(3cos(sin1(2x+3)))

3*d/dxcos(sin^-1(2x+3))3ddxcos(sin1(2x+3))

Use the chain rule:

3*d/(du)cos(u)*d/dxsin^-1(2x+3)3dducos(u)ddxsin1(2x+3)

3*d/(du)cos(u)*d/(dw)sin^-1(w)*d/dx(2x+3)3dducos(u)ddwsin1(w)ddx(2x+3)

3*(-sin(u))*1/sqrt(1-w^2)*23(sin(u))11w22

-6*sin(u)*1/sqrt(1-w^2)6sin(u)11w2

As u=sin^-1(2x+3)u=sin1(2x+3) and w=2x+3w=2x+3, we get:

-6*sin(sin^-1(2x+3))*1/sqrt(1-(2x+3)^2)6sin(sin1(2x+3))11(2x+3)2

-6*(2x+3)*1/sqrt(-4x^2+12x+10)6(2x+3)14x2+12x+10

-(12x+18)/(isqrt(4x^2-12x-10))12x+18i4x212x10

Mar 3, 2018

1) 12/sqrt( 1 - 9x^2)1219x2

2)(5+ 1/sqrt(1-x^2))/(5x + sin^-1 x)5+11x25x+sin1x

3) -(6 (2x + 3))/ sqrt(1 - (2x + 3)^2)6(2x+3)1(2x+3)2

Explanation:

You can Solve this Using the Chain Rule
basically,

d/dx f(g(x))ddxf(g(x))

= f'(g(x))* g'(x)

which is the derivative of the first function(f(x)) evaluated at g(x)
multiplied by the derivative of g(x)

therefore for the first question,
let f(x) = 4sin^-1(x)
and g(x) = 3x

therefore, for the derivative of f(g(x))

derivative of f(x) is 4* 1/sqrt(1-x^2)
and of g(x) Is 3

therefore , the total derivative is

4 * 1/sqrt(1 - (3x)^2) * 3
= 12 /sqrt(1 - 9x^2)

2)
similarly, for the second one,
derivative of ln(x) is 1/x which evaluated at the inside function is 1/ (5x + sin^-1(x)) multiplied by the derivative of the inside function of
5x + sin^-1(x)
is 5 + 1/sqrt(1 - x^2)

therefore the total derivative is
(5 + 1/sqrt(1 - x^2)) / (5x + sin^-1(x))

3)
dy/dx =d/dx 3 cos(sin^-1(2x+3))

use the same chain rule twice.
the derivative of the outside function = -3sin(x)
which has to be evaluated at the inside function,
sin^-1(2x+3)

And the derivative of the inside function found using the chain rule is
2* 1/ sqrt(1 - (2x + 3)^2)

therefore,
-3sin(sin^-1(2x+3)) *2 / sqrt(1 - (2x + 3)^2)

but since sin(sin^-1(2x+3)) = 2x+ 3
the entire derivative is

-(6(2x+3)) / sqrt(1 - (2x + 3)^2)

And That's it