find the derivative of (f(x)=sec(1 + x^{2})).\n(f(x)=sec(1 + x^{2})\tan(1 + x^{2})2x)\n(f(x)=-csc(1 +…

Question 1

Explanation:

Step1: Apply chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Let $u = 1 + x^{2}$, and $y=\sec(u)$. The derivative of $\sec(u)$ with respect to $u$ is $\sec(u)\tan(u)$, and the derivative of $u = 1 + x^{2}$ with respect to $x$ is $2x$.

Step2: Calculate the derivative

By the chain - rule, $f'(x)=\frac{d}{dx}\sec(1 + x^{2})=\sec(1 + x^{2})\tan(1 + x^{2})\cdot\frac{d}{dx}(1 + x^{2})=\sec(1 + x^{2})\tan(1 + x^{2})\cdot2x$

Answer:

$f'(x)=\sec(1 + x^{2})\tan(1 + x^{2})2x$

Question 2

Explanation:

Step1: Apply chain - rule

Let $u=\sqrt{x^{2}+\sin x}=(x^{2}+\sin x)^{\frac{1}{2}}$, and $y = e^{u}$. The derivative of $y = e^{u}$ with respect to $u$ is $e^{u}$, and the derivative of $u=(x^{2}+\sin x)^{\frac{1}{2}}$ with respect to $x$ can be found using the chain - rule again. Let $v=x^{2}+\sin x$, then $u = v^{\frac{1}{2}}$. The derivative of $u$ with respect to $v$ is $\frac{1}{2}v^{-\frac{1}{2}}$, and the derivative of $v$ with respect to $x$ is $2x+\cos x$. So $\frac{du}{dx}=\frac{1}{2}(x^{2}+\sin x)^{-\frac{1}{2}}(2x + \cos x)$.

Step2: Calculate the derivative

By the chain - rule, $f'(x)=\frac{d}{dx}e^{\sqrt{x^{2}+\sin x}}=e^{\sqrt{x^{2}+\sin x}}\cdot\frac{d}{dx}\sqrt{x^{2}+\sin x}=e^{\sqrt{x^{2}+\sin x}}\frac{1}{2}(x^{2}+\sin x)^{-\frac{1}{2}}(2x + \cos x)$

Answer:

$f'(x)=e^{\sqrt{x^{2}+\sin x}}\frac{1}{2}(x^{2}+\sin x)^{-\frac{1}{2}}(2x + \cos x)$