y = 3x^2 + 2x^(1/2) - 12 Find dy/dx
Firstly we divide up the equations into its three compenents based on the powers of the x values, giving us 3x^2, 2x^(1/2) and -12. Now one at a time, we multiply the coefficient by the power of x, and then subtract one from this power. For each component we get:
Multiply by power (2): 3x^2 * 2 = 6x^2 Subtract 1 from power: 2-1=1 --> 6x
Multiply by power (1/2): 2x^(1/2) * 1/2 = x^(1/2) Subtract 1 from power: (1/2 - 1 = -1/2) --> x^(-1/2) or 1/x^(1/2) as power is negative.
With the final component we can save our selves time by knowing that if you differentiate any number that isnt multipled by a variable, in this case x, it simply equals zero. This is because we view the number as being a coefficient of x^0, hence our first step is multiplying by zero.
The final answer is : dy/dx = 6x + 1/x^(1/2)
Related Maths A Level answers
All answers ▸Solve the simultaneous equations: y + 4x + 1 = 0, and y^2 + 5x^2 + 2x = 0.
How can I find the derivative of y = tan(x)?
The circle C has centre (3, 1) and passes through the point P(8, 3). (a) Find an equation for C. (b) Find an equation for the tangent to C at P, giving your answer in the form ax + by + c = 0 , where a, b and c are integers.
Simplify and solve for x. log(x+1)+log 5=2. Note, log is the natural log in this case