If the function is : $$f(x) = (2x-b)^a$$ And it's derivative is: $$f'(x) = 24x^2-24x+6$$ Then find the value of $a$ and $b$
I tried by calculating the derivative of $f(x)$ which comes out to be: $$f'(x) = 2a(2x-b)^{a-1}$$ And then by equating the two : $$24x^2-24x+6 = 2a(2x-b)^{a-1}$$
From now on I really don't know how to approach it, any kind of help would be appreciated .Thanks in advance.
Well, we compute:
$$\frac{\partial}{\partial x}\left(\left(\text{n}x-\text{b}\right)^\text{a}\right)=\text{a}\left(\text{n}x-\text{b}\right)^{\text{a}-1}\cdot\frac{\partial}{\partial x}\left(\text{n}x-\text{b}\right)=$$ $$\text{a}\left(\text{n}x-\text{b}\right)^{\text{a}-1}\cdot\left(\text{n}\cdot\frac{\text{d}}{\text{d}x}\left(x\right)-\text{b}\cdot\frac{\partial}{\partial x}\left(1\right)\right)=\text{a}\left(\text{n}x-\text{b}\right)^{\text{a}-1}\cdot\left(\text{n}\cdot1-\text{b}\cdot0\right)\tag1$$