Differentiation techniques 1. Use the first principle to find the derivative of each of the following functions func- tion at $$ x=4 $$. (a) $$ f(x)=x^{2}+1 $$ (b) $$ f(x)=x-6 $$ (c) $$ f(x)=x^{3}+2 $$ 2. Find $$ \frac{d y}{d x} $$ for each of the following functions (a) $$ y=x^{5}-\frac{3}{x^{2}}+8 x+1 $$ (b) $$ y=2 x-\frac{4}{\sqrt{x}} $$ (c) $$ y=\sin (5 x)+\ln x-\sqrt{x^{3}} $$ 3. Find $$ f^{\prime}(x) $$ for each of the following functions (a) $$ y=(x+3)^{7} $$ (b) $$ y=\cos (\sin x) $$ (c) $$ y=\ln \left(2 x^{3}-5\right) $$ (d) $$ y=\sin x $$ (a)

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### 1. Using the first principle at $$ x=4 $$

The definition of the derivative (first principle) is  
$$
f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}.
$$

#### (a) $$ f(x)=x^2+1 $$

1. Compute $$ f(4+h) $$:
   $$
   f(4+h)=(4+h)^2+1=16+8h+h^2+1=17+8h+h^2.
   $$
2. Compute $$ f(4) $$:
   $$
   f(4)=4^2+1=16+1=17.
   $$
3. Form the difference quotient:
   $$
   \frac{f(4+h)-f(4)}{h}=\frac{(17+8h+h^2)-17}{h}=\frac{8h+h^2}{h}=8+h.
   $$
4. Take the limit as $$ h\to0 $$:
   $$
   f'(4)=\lim_{h\to0}(8+h)=8.
   $$

#### (b) $$ f(x)=x-6 $$

1. Compute $$ f(4+h) $$:
   $$
   f(4+h)=(4+h)-6=4-6+h=-2+h.
   $$
2. Compute $$ f(4) $$:
   $$
   f(4)=4-6=-2.
   $$
3. Form the difference quotient:
   $$
   \frac{f(4+h)-f(4)}{h}=\frac{(-2+h)-(-2)}{h}=\frac{h}{h}=1.
   $$
4. Thus,
   $$
   f'(4)=1.
   $$

#### (c) $$ f(x)=x^3+2 $$

1. Compute $$ f(4+h) $$:
   $$
   f(4+h)=(4+h)^3+2=64+3\cdot16\,h+3\cdot4\,h^2+h^3+2=64+48h+12h^2+h^3+2=66+48h+12h^2+h^3.
   $$
2. Compute $$ f(4) $$:
   $$
   f(4)=4^3+2=64+2=66.
   $$
3. Form the difference quotient:
   $$
   \frac{f(4+h)-f(4)}{h}=\frac{(66+48h+12h^2+h^3)-66}{h}=\frac{48h+12h^2+h^3}{h}=48+12h+h^2.
   $$
4. Take the limit as $$ h\to0 $$:
   $$
   f'(4)=\lim_{h\to0}(48+12h+h^2)=48.
   $$

---

### 2. Differentiation using standard rules

#### (a) $$ y=x^5-\frac{3}{x^2}+8x+1 $$

Rewrite $$-\frac{3}{x^2}$$ as $$-3x^{-2}$$. Differentiating term by term:
- $$\frac{d}{dx}(x^5)=5x^4$$,
- $$\frac{d}{dx}(-3x^{-2})=-3(-2)x^{-3}=6x^{-3}$$,
- $$\frac{d}{dx}(8x)=8$$,
- $$\frac{d}{dx}(1)=0$$.

Thus, 
$$
y'=5x^4+6x^{-3}+8.
$$

#### (b) $$ y=2x-\frac{4}{\sqrt{x}} $$

Rewrite $$-\frac{4}{\sqrt{x}}$$ as $$-4x^{-1/2}$$. Differentiating:
- $$\frac{d}{dx}(2x)=2$$,
- $$\frac{d}{dx}(-4x^{-1/2})=-4\left(-\frac{1}{2}\right)x^{-3/2}=2x^{-3/2}$$.

Thus,
$$
y'=2+2x^{-3/2}.
$$

#### (c) $$ y=\sin(5x)+\ln x-\sqrt{x^3} $$

Rewrite $$\sqrt{x^3}$$ as $$ x^{3/2} $$. Differentiating:
- $$\frac{d}{dx}(\sin(5x))=5\cos(5x)$$ (using the chain rule),
- $$\frac{d}{dx}(\ln x)=\frac{1}{x}$$,
- $$\frac{d}{dx}(x^{3/2})=\frac{3}{2}x^{1/2}$$.

Thus,
$$
y'=5\cos(5x)+\frac{1}{x}-\frac{3}{2}x^{1/2}.
$$

---

### 3. Finding $$ f'(x) $$ for the given functions

#### (a) $$ y=(x+3)^7 $$

Apply the chain rule:
$$
y'=7(x+3)^{6}\cdot \frac{d}{dx}(x+3)=7(x+3)^6.
$$

#### (b) $$ y=\cos(\sin x) $$

Apply the chain rule twice:
$$
y'=-\sin(\sin x)\cdot \frac{d}{dx}(\sin x)=-\sin(\sin x)\cos x.
$$

#### (c) $$ y=\ln(2x^3-5) $$

Differentiate using the chain rule:
$$
y'=\frac{1}{2x^3-5}\cdot \frac{d}{dx}(2x^3-5)=\frac{1}{2x^3-5}\cdot 6x^2=\frac{6x^2}{2x^3-5}.
$$

#### (d) $$ y=\sin x $$

The derivative is:
$$
y'=\cos x.
$$
1. **Using the First Principle at $$ x=4 $$:** - **(a) $$ f(x) = x^2 + 1 $$** - $$ f'(4) = 8 $$ - **(b) $$ f(x) = x - 6 $$** - $$ f'(4) = 1 $$ - **(c) $$ f(x) = x^3 + 2 $$** - $$ f'(4) = 48 $$ 2. **Differentiation Using Standard Rules:** - **(a) $$ y = x^5 - \frac{3}{x^2} + 8x + 1 $$** - $$ y' = 5x^4 + 6x^{-3} + 8 $$ - **(b) $$ y = 2x - \frac{4}{\sqrt{x}} $$** - $$ y' = 2 + 2x^{-3/2} $$ - **(c) $$ y = \sin(5x) + \ln x - \sqrt{x^3} $$** - $$ y' = 5\cos(5x) + \frac{1}{x} - \frac{3}{2}x^{1/2} $$ 3. **Finding $$ f'(x) $$ for the Given Functions:** - **(a) $$ y = (x + 3)^7 $$** - $$ f'(x) = 7(x + 3)^6 $$ - **(b) $$ y = \cos(\sin x) $$** - $$ f'(x) = -\sin(\sin x)\cos x $$ - **(c) $$ y = \ln(2x^3 - 5) $$** - $$ f'(x) = \frac{6x^2}{2x^3 - 5} $$ - **(d) $$ y = \sin x $$** - $$ f'(x) = \cos x $$

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