Basic differential calculus problems

We apply the Sum rule and Power rule: \begin{eqnarray} \dfrac{df(x)}{dx} &=& \dfrac{d}{dx}(3x^2 + 7) \\ \ &=& 3 \dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(7) \\ \ &=& 3\, (2x) + 0 \\ \ &=& 6x. \end{eqnarray}
Actually, almost no one expects you to explicitly apply the Sum rule. It is usually more than acceptable to leave that step out: \begin{eqnarray} f'(x) &=& (3x^2 + 7)' \\ \ &=& 3\, (2x) + 0 \\ \ &=& 6x. \end{eqnarray}

First, recall that $$ (x^3)' = 3x^2$$ and $$(\sin x)' = \cos x.$$ Then, we apply the Product rule: \begin{eqnarray} f'(x) &=& (x^3 \sin (x))' \\ \ &=& (x^3)' \sin (x) + x^3 (\sin (x))' \\ \ &=& 3x^2\sin (x) + x^3 \cos (x). \end{eqnarray}

First, recall that the tangent function is given by $$\tan x = \dfrac{\sin x}{\cos x} = (\sin x)\, (\cos x)^{-1}$$ and that the derivatives of $\sin x$ and $\cos x$ are $$\dfrac{d(\sin x)}{dx} = \cos x$$ and $$\dfrac{d(\cos x)}{dx} = -\sin x.$$ Next, we apply the Product rule and Chain rule: \begin{eqnarray} \dfrac{df(x)}{dx} &=& \dfrac{d}{dx} \left[(\sin x)\, (\cos x)^{-1}\right] \\ \ &=& \dfrac{d(\sin x)}{dx}\, (\cos x)^{-1} + (\sin x)\dfrac{d(\cos x)^{-1}}{dx} \\ \ &=& (\cos x)\, (\cos x)^{-1} + (\sin x) [-(\cos x)^{-2} \dfrac{d(\cos x)}{dx} ] \\ \ &=& 1 + (\sin x)\, [-(\cos x)^{-2} (-\sin x)] \\ \ &=& 1 + (\sin x)^2 \, (\cos x)^{-2} \\ \ &=& 1 + \tan^2 x \\ \ &=& \sec^2 x. \end{eqnarray}

First, recall that $$(\sin x)' = \cos x$$ and $$(x^{-1})' = -x^{-2}.$$ Next, we rewrite $f(x)$ as $$f(x) = (4x^3 - 2x)\, (\sin x)^{-1}$$ and apply to it the Product rule and Chain rule: \begin{eqnarray} f'(x) &=& [(4x^3-2x)\, (\sin x)^{-1}]' \\ \ &=& (4x^3 - 2x)'\, (\sin x)^{-1} + (4x^3-2x)\, [(\sin x)^{-1}]' \\ \ &=& \dfrac{12x^2 - 2}{\sin x} + (4x^3-2x)\, [-(\sin x)^{-2}(\sin x)'] \\ \ &=& \dfrac{12x^2 - 2}{\sin x} + (4x^3-2x)\, [-(\sin x)^{-2}(\cos x)] \\ \ &=& \dfrac{12x^2 - 2}{\sin x} - \dfrac{(4x^3-2x)\, \cos x}{\sin^2 x}. \end{eqnarray}

First, recall that $$(e^x)' = e^x.$$ Next, we apply the Chain rule: \begin{eqnarray} f'(x) &=& \exp(4x^7-5)\, (4x^7 - 5)' \\ \ &=& \exp(4x^7-5)\, (28x^6) \\ \ &=& 28x^6 \exp(4x^7-5). \end{eqnarray} $$ $$ The last step is not actually needed. It just looks better this way is all.

First, recall that $$(\cos x)' = -\sin x.$$ Next, we repeatedly apply the Chain rule: \begin{eqnarray} f'(x) &=& [\left(x^3 - 9\cos(3x)\right)^5]' \\ \ &=& 5(x^3 - 9\cos(3x))^4 [x^3 - 9\cos(3x)]' \\ \ &=& 5(x^3 - 9\cos(3x))^4 (3x^2 - 9 (\cos (3x))') \\ \ &=& 5(x^3 - 9\cos(3x))^4 (3x^2 - 9 (- \sin (3x))(3x)') \\ \ &=& 5(x^3 - 9\cos(3x))^4 (3x^2 + 27 \sin (3x)). \end{eqnarray}

First, we rewrite $f(x)$ as $$f(x) = x^{1/2}.$$ Next, we apply the Power rule: \begin{eqnarray} f'(x) &=& (x^{1/2})' \\ \ &=& \dfrac{1}{2}x^{1/2 -1} \\ \ &=& \dfrac{1}{2}x^{-1/2} \\ \ &=& \dfrac{1}{2\sqrt{x}}. \end{eqnarray}
Note: It is perfectly fine leaving the answer as $$f'(x) = \dfrac{1}{2}x^{-1/2}.$$

Your first thought might be to apply the Power rule to this, but no. The Power rule is only for when the exponent is a constant. The exponent here is a variable so the Power rule does not apply.

Instead, we rewrite $f(x)$ as \begin{eqnarray} f(x) &=& \exp( \ln (3^x)) \\ \ &=& \exp (x \ln 3). \end{eqnarray} Then, we recall that $$(\exp x)' = \exp$$ and apply the Chain rule: \begin{eqnarray} f'(x) &=& \exp(x \ln 3)\, (x \ln 3)' \\ \ &=& \exp(x \ln 3) \, (\ln 3) \\ \ &=& 3^x (\ln 3). \end{eqnarray}