$f’((\sin x)^2)=(\cos x)^2$. Then is $~f(x) =(\cos x)^2+c$?

Question

$f'((\sin x)^2)=(\cos x)^2$. Then is $f(x) =(\cos x)^2+c \text{ ?}$

First I don’t know what can $f’((\sin x)^2)$ be presented in the $\dfrac{d\cdots}{d\cdots}$ form

Then $(\sin x)^2$ in the $f'$ confused me. Since $f'((\sin x)^2)=1-(\sin x)^2$, can I just simply substitute $(\sin x)^2$ with $x$ straight ?

Answer 1

$f'$ is basically the derivative of $f$ with respect to the single variable that it accepts as an argument. I'm sure you have no issue with understanding the meaning of $f'(w)$ when you are explicitly given $f(w)$ in terms of $w$. But the situation can become a bit less clear when you introduce other variables in place of $w$. In these cases, you should view $f$ as just a formal functional definition acting on a single variable (such as $w$), find $f'$ first by differentiating with respect to $w$, then substitute for $w$ with the expression as you're given only after differentiation.

For example, if $f(w) = w^2 + 3w + 1$, then $f'(w) = 2w+3$ and $f'(2u^5 - 1) = 2(2u^5-1) + 3$.

When you're presented with $f'(\sin^2 x) = \cos^2x$, you should try to work on the argument to make it into a simple, uncomplicated variable that can easily be related to the right hand side. This is not always possible, but it is easy here. Use the identity $\sin^2x + \cos^2x = 1$ to rewrite the above as:

$f'(1-\cos^2x) = \cos^2x$

and now make the substitution $y = 1-\cos^2x$ to give:

$f'(y) = 1-y$

You can now write:

$\frac{d}{dy}f(y) = 1-y$, and solve this simple differential equation by separation of variables to give:

$f(y) = y - \frac 12y^2 + c$

And now we can simply write $f(x) = x - \frac 12x^2 + c$

since all we need to do is replace $y$ with $x$ in the formal definition of $f$ that we've found.

Answer 2

$\cos(x)^2 = 1 - \sin(x)^2$, so your equation is saying $f'(t) = 1 - t$ for $0 \le t \le 1$ (the range of values of $\sin(x)^2$, assuming you're only dealing with real numbers).