# `f(x) = 10x^10 + 5x^5 - x` Find the first and second derivatives of the function.

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`f'(x)=d/dx(10x^10 +5x^5-x)`

`f'(x)=d/dx(10x^10)+d/dx(5x^5)-d/dx(x)`

`f'(x)=100x^9+25x^4-1`

`f''(x)=d/dx(100x^9)+d/dx(25x^4)-d/dx(1)`

`f''(x)=900x^8+100x^3`

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Given f(x) = 10(x^10) + 5(x^5) - x

Derive on both sides with respect to x

f'(x) = 10[10(x^9)] + 5[5(x^4)] -1

= 100(x^9) + 25(x^4) -1

Now again derive on both sides with respect to x to get second derivative of x

f''(x) = 100[9(x^8)] + 25[4(x^3)] - 0

= 900(x^8) + 100(x^3)

Thus we can get the first and second derivatives of x.

Q.) f(x) = 10x^10 + 5x^5 - x

**Solution:- **

**NOTE: 1) If f(x) = x^n ; where n = integer; then **

**f'(x) = d[f(x)]/dx = n*x^(n-1) **

**2) If f(x) = k ; where 'k' = real number, then**

**f'(x) = d[f(x)]/dx = 0**

Hence, f(x) = 10x^10 + 5x^5 - x

Differentiating once with respect to 'x' we get,

f'(x) = d[f(x)]/dx = (10*10)*(x^9) + (5*5)*(x^4) - 1*(x^0)

or, f'(x) = 100*(x^9) + 25*(x^4) - 1 ; (since x^0 = 1)..................(A)

Again differentiating f'(x) with respect to 'x' we get

f"(x) = d[f'(x)]/dx = (100*9)*(x^8) + (25*4)*(x^3) - 0

or, f"(x) = 900*(x^8) + 100*(x^3)

or, f'(x) = (100*x^3)*[9*(x^5) + 1]

f'(x)=100x^9 + 25x^4-1

f''(x)=900x^8+100x^3

f(x) = 10x10 + 5x5 – x

AxB

In the example term above, A is the coefficient and B is the exponent.

To find the derivative, you take the exponent and multiply that number by the current coefficient. The result of this becomes the new coefficient. Then, the exponent decreases by a value of 1. I will break down the first term of the function.

10x10

The coefficient is 10, and the exponent is 10.

Step 1: Multiply the coefficient by the exponent.

10 x 10 = 100

Step 2: That result becomes the new coefficient.

100x?

Step 3: The exponent decreases by a value of 1.

10-1 = 9

Therefore, the derivative of the first term, 10x10, is 100x9

Following those steps for all terms, the first derivative of the function is

f’(x) = 100x9 + 25x4 – 1

The last term may be a little tricky. It starts at 1x1. 1 x 1 = 1, so that is the new coefficient. Then, the exponent gets decreased by a value of 1, so it goes from 1 to 0. Any number raised to the 0 power is 1, which leaves 1 as our final term in the first derivative.

The 2nd derivative follows the same steps, leaving:

f''(x) = 900x8 + 100x3

`f(x)=10x^(10)+5x^(5)-x`

`f'(x)=10(10)x^(10-1)+5(5)x^(5-1)-1`

`=100x^9+25x^4-1`

`f'(x)=100x^9+25x^4-1`

`f''(x)=100(9)x^(9-1)+25(4)x^(4-1)-0`

`f''(x)=900x^8+100x^3`

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