Differentiation questions

Questions

• 1. Differentiate x^2(2-3x)+(x^2+1/x)
• 2. Differentiate 1/x^3 - 2/x^5
• 3. Differentiate root(x)+1/(cube root x)
• 4. Differentiate x^2(1/(cube root x) -2x^-5)
• 5. Differentiate x(root(x) + 1)(1/x -2) and find the gradient at x=1.

Answers 1,2 and 3

• 1. Rewrite as f(x) = 2x^2-3x^3+x^2/x + 1/x , and then again in terms of the powers of x, so: f(x) = 2x^2-3x^3 + x + x^-1
• Now we can differentiate this to get f'(x) = 2(2x) - 3(3x^2) + 1(x^0) + (-1)x^-2 , which can be simplified to get f'(x) = 4x-9x^2+1-x^-2. This is the same as 4x-9x^3+1- 1/x^2
• 2. Rewrite as f(x) = x^-3 - 2x^-5 and differentiate to get f'(x) = -3x^-4 + 10x^-6, which is the same as -3/x^4 + 10/x^6
• 3. Rewrite as f(x) = x^(1/2) + x^(-1/3) and differentiate to get f'(x)= 1/2x^(-1/2) - 1/3x^(-4/3) , which can be rewritten as 1/2 x 1/x^(1/2) - 1/3 x 1/x^(4/3) which is the same as 1/2(root x) - 1/3(cube root x)^4

Answers 4 and 5

• 4. Rewrite as x^2(1/3x^(-1/2) - 2x^-5) and then multiply by the x^2 outside the bracket to get f(x)= 1/3x^(3/2) - 2x^-3. Differentiate this to get 1/2x^(1/2) + 6x^-4. This is the same as root(x)/2 + 6/x^4
• 5. Rewrite as x(x^(1/2)+1)(x^-1 -2) , then multiply by whats outside the brackets to give f(x)= (x^(3/2)+x)(x-1 -3). This can be expanded to give x^1/2 - 2x^(3/2) + 1 - 2x, which can be differentiated to give 1/2x^(-1/2) - 3x^(1/2) - 2 , which is the same as 1/2(root x) - 3(root x) -2.
• Now we do dy/dx (x=1) which is 1/2 - 3 -2 which equals the graident at x=1 ; -9/2.