Question

for full credit, please show all steps clearly. find the derivatives of the given functions. a) f(x)=sin(x^2 - 3x + 4) f(x)= b) y = tan^3(x) y= c) f(x)=cos^4(x^2 - 3x + 5) df(x)/dx = d) y=(x^2 - 3x + 5)cos^4(x) dy/dx =

Explanation:

Step1: Apply chain - rule for a)

Let $u = x^{2}-3x + 4$, then $f(x)=\sin(u)$. The chain - rule states that $f^\prime(x)=\frac{df}{du}\cdot\frac{du}{dx}$. We know that $\frac{d}{du}\sin(u)=\cos(u)$ and $\frac{d}{dx}(x^{2}-3x + 4)=2x - 3$. So $f^\prime(x)=\cos(x^{2}-3x + 4)\cdot(2x - 3)$.

Step2: Apply chain - rule for b)

Let $u=\tan(x)$, then $y = u^{3}$. By the chain - rule $y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}$. We know that $\frac{d}{du}u^{3}=3u^{2}$ and $\frac{d}{dx}\tan(x)=\sec^{2}(x)$. So $y^\prime=3\tan^{2}(x)\sec^{2}(x)$.

Step3: Apply chain - rule for c)

Let $u = x^{2}-3x + 5$ and $v=\cos(u)$, then $f(x)=v^{4}$. By the chain - rule, $\frac{df}{dx}=\frac{df}{dv}\cdot\frac{dv}{du}\cdot\frac{du}{dx}$. We have $\frac{d}{dv}v^{4}=4v^{3}$, $\frac{d}{du}\cos(u)=-\sin(u)$ and $\frac{d}{dx}(x^{2}-3x + 5)=2x - 3$. So $f^\prime(x)=4\cos^{3}(x^{2}-3x + 5)\cdot(-\sin(x^{2}-3x + 5))\cdot(2x - 3)=-4(2x - 3)\cos^{3}(x^{2}-3x + 5)\sin(x^{2}-3x + 5)$.

Step4: Apply product - rule for d)

The product - rule states that if $y = g(x)h(x)$, then $y^\prime=g^\prime(x)h(x)+g(x)h^\prime(x)$. Let $g(x)=x^{2}-3x + 5$ and $h(x)=\cos^{4}(x)$. We know that $g^\prime(x)=2x - 3$. For $h(x)=\cos^{4}(x)$, let $u=\cos(x)$, then $h(x)=u^{4}$, and $\frac{dh}{dx}=\frac{dh}{du}\cdot\frac{du}{dx}=4u^{3}\cdot(-\sin(x))=-4\cos^{3}(x)\sin(x)$. So $y^\prime=(2x - 3)\cos^{4}(x)+(x^{2}-3x + 5)\cdot(-4\cos^{3}(x)\sin(x))=(2x - 3)\cos^{4}(x)-4(x^{2}-3x + 5)\cos^{3}(x)\sin(x)$.

Answer:

a) $f^\prime(x)=(2x - 3)\cos(x^{2}-3x + 4)$
b) $y^\prime=3\tan^{2}(x)\sec^{2}(x)$
c) $f^\prime(x)=-4(2x - 3)\cos^{3}(x^{2}-3x + 5)\sin(x^{2}-3x + 5)$
d) $y^\prime=(2x - 3)\cos^{4}(x)-4(x^{2}-3x + 5)\cos^{3}(x)\sin(x)$