Find d^5/dx^5 g(x) for g(x)=sinx+5x^4

A. cos x
B. sin x + 120
C. -cos x +20x^3
D. -cos x + 120 x
E. sin x

On Apex I found the answer to cos(x)

d^4/dx^4 sinx = sinx

The n+1st derivative of x^n is zero
so, A

To find the fifth derivative of g(x) = sin(x) + 5x^4 with respect to x, we can differentiate each term of the function separately and then multiply by the appropriate coefficients.

The derivatives of the terms are as follows:

d/dx (sin(x)) = cos(x)
d/dx (5x^4) = 20x^3

Since we want the fifth derivative, we need to differentiate each term four more times. Differentiating cos(x) four more times gives us:

d^5/dx^5 (cos(x)) = -cos(x)

Differentiating 20x^3 four more times gives us:

d^5/dx^5 (20x^3) = 120

Now we can put everything together:

d^5/dx^5 (g(x)) = d^5/dx^5 (sin(x) + 5x^4)
= d^5/dx^5 (sin(x)) + d^5/dx^5 (5x^4)
= -cos(x) + 120

Therefore, the fifth derivative of g(x) is -cos(x) + 120.

The correct answer is option D.

To find the fifth derivative of g(x), you would need to differentiate g(x) five times. Let's break it down step by step.

Step 1: Find the first derivative of g(x):
g'(x) = d/dx (sinx + 5x^4)

The derivative of sinx is cosx, and the derivative of 5x^4 is 20x^3 (using the power rule). Therefore, the first derivative of g(x) is:

g'(x) = cosx + 20x^3

Step 2: Find the second derivative of g(x):
g''(x) = d/dx (g'(x))

The derivative of cosx is -sinx, and the derivative of 20x^3 is 60x^2 (using the power rule). Therefore, the second derivative of g(x) is:

g''(x) = -sinx + 60x^2

Step 3: Find the third derivative of g(x):
g'''(x) = d/dx (g''(x))

The derivative of -sinx is -cosx, and the derivative of 60x^2 is 120x (using the power rule). Therefore, the third derivative of g(x) is:

g'''(x) = -cosx + 120x

Step 4: Find the fourth derivative of g(x):
g''''(x) = d/dx (g'''(x))

The derivative of -cosx is sinx, and the derivative of 120x is 120 (since the derivative of a constant is zero). Therefore, the fourth derivative of g(x) is:

g''''(x) = sinx + 120

Step 5: Find the fifth derivative of g(x):
g'''''(x) = d/dx (g''''(x))

The derivative of sinx remains sinx, and the derivative of 120 is zero (since it is a constant). Therefore, the fifth derivative of g(x) is:

g'''''(x) = sinx

So, the fifth derivative of g(x) is sinx.

Therefore, the correct answer is A. sin x