Find y^(20)(0) of y=(x+3)cos(2x^3)

Question

y=(x+3)cos(2x³)

Find y⁽²⁰⁾(0).

I need explain!

Answer 1

To find the 20th derivative of the function y=(x+3)cos⁡(2x³), we can use the Leibniz rule for differentiation, which allows us to differentiate the product of two functions.

The n-th derivative of a product of two functions f(x) and g(x) is given by:

(fg)^(n)(x)= \sum_{k=0}^{n} \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x)(fg)(n)(x)

In our case, let:

1. f(x)=x+3
2. g(x)=cos⁡(2x³)

Step 1: Compute the Derivatives

Derivative of f(x)=x+3

The derivatives of f(x)f

1. f(x)=x+3
2. f^{(0)}(x) = x + 3
3. f\(x)=1
4. f^k(x)=0 for k≥2

Derivative of g(x)=cos⁡(2x³)

We need to compute the derivatives of g(x) using the chain rule. The derivatives will involve the function cos⁡\cos and the derivative of 2x^3

Using the chain rule, we have:

g′(x)== -12x^2 \sin(2x^3)

To compute higher derivatives, we observe that:

1. The even derivatives of g(x) will involve cos(2x^3) evaluated at 0.
2. The odd derivatives will involve sin(2x^3) evaluated at 0.

At x=0

1. g(0)=cos⁡(0)=1
2. g′(0)=−12

Step 2: Determine Derivatives at x=0

For k=0

g^{(20)}(0) = 20th derivative of g(x) evaluated at 0

We need to compute g^{(20) Since every derivative of g(x) that is an odd derivative equals zero at x=0 (since all odd derivatives will involve sin(0), we need to focus only on even derivatives.

The pattern generally involves:

1. Even derivatives will contribute to cos(0)
2. Odd derivatives will be 0 at x=0

Contribution to y^(20)(0)

Using the Leibniz rule:

y^{(20)}(0) = \sum_{k=0}^{20} \binom{20}{k} f^{(k)}(0) g^{(20-k)}(0)y(20)(0)

The only non-zero contributions come from k=0 (which gives g(20) and k=1 (which gives f′(0)⋅g(19)(0)f'(0) \cdot g^{(19)}(0)f′(0)⋅g(19)(0), but g(19)(0)=0

.

Thus,

y^{(20)}(0) = 1 *g^{(20)}(0)

Now we need to find g^{(20)}(0). The pattern of derivatives involves increasingly complex polynomials multiplied by cos(2x^3 and sin(2x^3). However, without calculating all previous derivatives of g, we can conclude that:

y^(20)(0)=0, as the contributions from odd derivatives vanish.

Thus,y^{(20)}(0) = 0.