Differentiate y = (x^2 + 1)^1/3
Use the chain rule to do this. First set u= x^2 + 1. We chose u to be this because u1/3 is much simpler to differentiate. Then find du/dx = 2x. Now find dy/du = 1/3 * u-2/3 = 1/3 * (x2 +1)-2/3. Now by the chain rule, dy/dx = dy/du * du/dx. Therefore dy/ dx = 2x * (1/3 * (x2 +1)-2/3 )= 2x/3 * (x2+1)-2/3
WW
Related Maths A Level answers
All answers ▸The second and fourth term of a geometric series is 100 and 225 respectively. Find the common ratio and first term of the series. Round your answer to 2 d.p if necessary
A curve C has equation: y = x^2 − 2x − 24x^1/2, x > 0; Find (i) dy/dx (ii) d^2y/dx^2
I don't understand why the function "f(x)=x^2 for all real values of x" has no inverse. Isn't sqrt(x) the inverse?
∫2x(x+2)^(1/2) dx evaluated from 0->2