f(x) = sin(x^2 + 3) . Innere Ableitung mal äussere Ableitung

Question

Kettenregel: f'(x) = u'(v(x)) * v'(x)

1) sin(x²+3) : v(x)=

                        v'(x)=

                        u'(x)=

                        u'(v(x))=

2) f(x) = (2x+9)⁴

3) f(x) = (x²+1)^1/2

4) 3cos(3x) = f(x)

Answer 1

f(x) = (2x+9)^4

4*(2x+9)^3*2


f(x) = (x^2+1)1/2

=1/2(x^2+1)^-1/2*2x


:)

Answer 2

Hi kassiopeoa

1) sin(x²+3) : v(x)= x^2+3

                        v'(x)= 2x

                        u'(v)= cos(v)

                        u'(v(x))= cos(x^2+3)

f'(x) = 2x*cos(x^2+3)

 

2) f(x) = (2x+9)⁴

f'(x) = 4*(2x+9)^3*2 = 8(2x+9)^3

 

3) f(x) = (x²+1)^1/2

f'(x) = 1/2*(x^2+1)^{-1/2}*2x = x*(x^2+1)^{-1/2}

 

4) 3cos(3x) = f(x)

f'(x) = 3*(-sin(3x))*3 = -9sin(3x)

 

Alles klar?

Grüße

Answer 3

1) f(x) = sin(x^2 + 3)

u(x) = sin(x)
v(x) = x^2 + 3

f'(x) = cos(x^2 + 3) * 2x

 

2) f(x) = (2x + 9)^4

u(x) = x^4
v(x) = 2x + 9

f'(x) = 4 * (2x + 9)^3 * 2 = 8 * (2x + 9)^3

 

3) f(x) = (x^2 + 1)^{1/2}

u(x) = x^{1/2}
v(x) = x^2 + 1

f'(x) = (1/2)*(x^2 + 1)^{-1/2} * 2x = x / (x^2 + 1)^{1/2}

 

4) f(x) = 3 * cos(3x)

u(x) = 3 * cos(x)
v(x) = 3x

f'(x) = 3 * (-sin(3x)) * 3 = -9 * sin(3x)

Answer 4

Hallo kassiopeia,

 

1) f(x) = sin(x²+3)

Innere Ableitung = 2x

Äußere Ableitung = cos(x² + 3)

f'(x) = 2x * cos(x² + 3)

 

2) f(x) = (2x+9)⁴

Innere Ableitung = 2

Äußere Ableitung = 4 * (2x + 9)³

f'(x) = 8 * (2x + 9)³

 

3) f(x) = (x²+1)^1/2

Innere Ableitung = 2x

Äußere Ableitung = 1/2 * (x² + 1)^-1/2

f'(x) = x * (x² + 1)^-1/2

 

4) f(x) = 3cos(3x)

Innere Ableitung = 3

Äußere Ableitung = -3 * sin(3x)

f'(x) = -9 * sin(3x)

 

Besten Gruß