Err, wrong. Partial derivative means derivative wrt the explicit variable **only**. What you took is the total derivative wrt $$\theta$$. Think about:

$$f(\theta,u,x)=3 \theta +u^2+ln(x)$$

where $$u=sin(\theta)$$.

I said

**as written**. There is no intermediate variable $$u$$ in the expression $$3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x)$$.

**As written** that's an expression that depends only on two independent variables, and its partial derivative with respect to $$\theta$$ is $$3 \,+\, \sin(2\theta)$$.

The way you define $$f$$ here is, incidentally,

*ambiguous*:

Example:

$$f=3 \theta+ sin^2(\theta)+ln(x)$$

$$\frac{df}{d \theta}=3+2 sin(\theta) cos (\theta)$$

$$\frac{\partial f}{\partial \theta}=3$$

On its own, that is just defining the function $$f(\theta,\,x)$$ to be equal to the expression above. There is no mention of $$u$$. The only hint that there should be such an intermediate variable at all is from earlier in your post, in the explanation that's supposed to motivate the example. But even there you never say how $$f$$ is supposed to depend on $$u$$. You may have

*meant*
$$

f(\theta,\, u,\, x) \,=\, 3\theta \,+\, u \,+\, \ln(x) \,,

$$

with $$u \,=\, \sin(\theta)^{2}$$, in which case $$\frac{\partial f}{\partial \theta} = 3$$. But you never said that. From what you wrote, $$f$$ could just as well have been

$$

f(\theta,\, u,\, x) \,=\, u \,+\, \sin(\theta)^{2} \,+\, \ln(x) \,,

$$

with $$u \,=\, 3\theta$$, or

$$

f(\theta,\, u,\, x) \,=\, \theta \,+\, u \,+\, \sin(\theta)^{2} \,+\, \ln(x) \,,

$$

with $$u \,=\, 2 \theta$$. In fact if you take your example literally and add that $$f$$ is supposed to depend on $$u$$, then your example just reads like

$$

f(\theta,\, u,\, x) \,=\, 3\theta \,+\, \sin(\theta)^{2} \,+\, \ln (x) \,,

$$

i.e. with $$\frac{\partial f}{\partial u} \,=\, 0$$.

Either way, the way you wrote it, saying just

$$f \,=\, 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \,,$$

anyone who sees that is going to tell you $$\frac{\partial f}{\partial \theta} \,=\, 3 \,+\, \sin(2\theta)$$.

And as others have pointed out, you certainly

*are* wrong to say that

$$

\frac{\mathrm{d}f}{\mathrm{d}\theta} \,=\, 3 \,+\, \sin(2\theta)

$$

because $$x$$ is still an independent parameter that needs to be kept constant. The derivative when $$\theta$$ is allowed to vary through $$u$$ is still a partial derivative, albeit a different one than when $$u$$ is kept constant. As I said in post #22, if you want to be technically correct you should really state explicitly that you're defining a new function $$f'(\theta,\, x) \,=\, f(\theta,\, u(\theta),\, x)$$ and then say you're taking the partial derivative $$\frac{\partial f'}{\partial \theta}$$, if you really need to make it unambiguous which partial derivative you're taking.