Solve this math calculus questions step by step Y=...

Below are full, line‑by‑line differentiations of the four functions. I keep the notation compact—$y$ is the dependent variable and $x$ the independent one.

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1. $y=\cos\bigl(nx\bigr)$

Chain rule with the outer function $\cos(u)$ and inner function $u = nx$.

$$\frac{dy}{dx} =\frac{d}{du}\bigl[\cos u\bigr]\;\frac{du}{dx} =(-\sin u)\,(n) =-n\sin\bigl(nx\bigr)$$

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2. $y=\sqrt{x^{2}+1}=(x^{2}+1)^{1/2}$

Rewrite with an exponent, then apply the chain rule.
Inner $u=x^{2}+1$, outer $u^{1/2}$.

$$\frac{dy}{dx} =\frac{1}{2}(u)^{-1/2}\,\frac{du}{dx} =\frac{1}{2}(x^{2}+1)^{-1/2}\,(2x) =\frac{x}{\sqrt{x^{2}+1}}$$

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3. $y=(2x+1)^{8}$

Again chain rule.
Inner $u=2x+1$, outer $u^{8}$.

$$\frac{dy}{dx} =8u^{7}\,\frac{du}{dx} =8(2x+1)^{7}\,(2) =16(2x+1)^{7}$$

(You could leave it at $16(2x+1)^{7}$ or expand further if required.)

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4. $y=e^{\sin x}$

Exponential‑of‑a‑function form.
Inner $u=\sin x$, outer $e^{u}$.

$$\frac{dy}{dx} =e^{u}\,\frac{du}{dx} =e^{\sin x}\,(\cos x) =e^{\sin x}\cos x$$

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Summary of the derivatives

$$\boxed{\; \begin{aligned} \frac{d}{dx}\bigl[\cos(nx)\bigr] &= -n\sin(nx)\\ \frac{d}{dx}\Bigl[\sqrt{x^{2}+1}\Bigr] &= \dfrac{x}{\sqrt{x^{2}+1}}\\ \frac{d}{dx}\bigl[(2x+1)^{8}\bigr] &= 16(2x+1)^{7}\\ \frac{d}{dx}\bigl[e^{\sin x}\bigr] &= e^{\sin x}\cos x \end{aligned} \;}$$

Each result follows directly from the chain rule, with power, trigonometric, and exponential derivative facts as needed.