Second order derivative of sin^4x

KeerthysrinivasanKS11

KeerthysrinivasanKS11

Answered question

2022-05-03

Second order derivative of sin^4x

Answer & Explanation

karton

karton

Expert2022-07-07Added 613 answers

Find the first derivative.

Differentiate using the chain rule, which states that ddx[f(g(x))]  is f(g(x))g(x) where f(x)=x4 and g(x)=sin(x).

f(x)=4sin3(x)ddx(sin(x))

The derivative of sin(x) with respect to x is cos(x).

f(x)=4sin3(x)cos(x)

 

Find the second derivative.

Since 4 is constant with respect to xx, the derivative of 4sin3(x)cos(x) with respect to x is 4ddx[sin3(x)cos(x)].

f′′(x)=4ddx(sin3(x)cos(x))

Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=sin3(x) and g(x)=cos(x).

f′′(x)=4(sin3(x)ddx(cos(x))+cos(x)ddx(sin3(x)))

The derivative of cos(x) with respect to x is -sin(x).

f′′(x)=4(sin3(x)(-sin(x))+cos(x)ddx(sin3(x)))

Multiply sin3(x) by sin(x) by adding the exponents.

f′′(x)=4(sin4(x)-1+cos(x)ddx(sin3(x)))

Simplify the expression.

f′′(x)=4(-sin4(x)+cos(x)ddx(sin3(x)))

Differentiate using the chain rule, which states that ddx[f(g(x))]  is  f(g(x))g(x) where f(x)=x3 and g(x)=sin(x).

f′′(x)=4(-sin4(x)+cos(x)(3sin2(x)ddx(sin(x))))

Move 3 to the left of cos(x).

f′′(x)=4(-sin4(x)+3(cos(x)sin2(x))ddx(sin(x)))

The derivative of sin(x) with respect to x is cos(x).

f′′(x)=4(-sin4(x)+3cos(x)sin2(x)cos(x))

Raise cos(x) to the power of 1.

f′′(x)=4(-sin4(x)+3(cos(x)cos(x))sin2(x))

Use the power rule aman=am+n to combine exponents.

f′′(x)=4(-sin4(x)+3cos(x)1+1sin2(x))

Add 1 and 1.

f′′(x)=4(-sin4(x)+3cos2(x)sin2(x))

Simplify.

f''(x)=12cos2(x)sin2(x)4sin4(x)

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