Find dy/dx by using chain rule Find dy/dx by using produce rule Find dy/dx by using Quotient. rule.

Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the differentiation problems using the specified rules.

1. Find $\frac{dy}{dx}$ by using the chain rule

a) $y = \sqrt{\sec x}$

Rewrite $y$ as $(\sec x)^{1/2}$ .
Let $u = \sec x$ . Then $y = u^{1/2}$ .

Differentiate $y$ with respect to $u$ and $u$ with respect to $x$ .

$$
\frac{dy}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}
$$
$$
\frac{du}{dx} = \sec x \tan x
$$

Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

$$
\frac{dy}{dx} = \frac{1}{2\sqrt{\sec x}} \cdot (\sec x \tan x) = \frac{\sec x \tan x}{2\sqrt{\sec x}}
$$
$$
\frac{dy}{dx} = \frac{\sqrt{\sec x} \tan x}{2}
$$
The final answer is $\boxed{\frac{\sqrt{\sec x} \tan x}{2}}$.

b) $y = \sqrt{4x^2 - 3x}$

Rewrite $y$ as $(4x^2 - 3x)^{1/2}$ .
Let $u = 4x^2 - 3x$ . Then $y = u^{1/2}$ .

Differentiate $y$ with respect to $u$ and $u$ with respect to $x$ .

$$
\frac{dy}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}
$$
$$
\frac{du}{dx} = 8x - 3
$$

Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .
$$ \frac{dy}{dx} = \frac{1}{2\sqrt{4x^2 - 3x}} \cdot (8x - 3) $$ The final answer is $\boxed{\frac{8x - 3}{2\sqrt{4x^2 - 3x}}}$.
- c) $y = \frac{1}{\cos^3 x}$
Rewrite $y$ as $(\cos x)^{-3}$ .
Let $u = \cos x$ . Then $y = u^{-3}$ .

Differentiate $y$ with respect to $u$ and $u$ with respect to $x$ .

$$
\frac{dy}{du} = -3u^{-4} = -\frac{3}{u^4}
$$
$$
\frac{du}{dx} = -\sin x
$$

Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

$$
\frac{dy}{dx} = -\frac{3}{(\cos x)^4} \cdot (-\sin x) = \frac{3\sin x}{\cos^4 x}
$$
$$
\frac{dy}{dx} = 3 \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\cos^3 x} = 3 \tan x \sec^3 x
$$
The final answer is $\boxed{3 \tan x \sec^3 x}$.

d) $y = \sqrt{\tan 3x}$

Rewrite $y$ as $(\tan 3x)^{1/2}$ .
Let $u = \tan 3x$ . Then $y = u^{1/2}$ .

Differentiate $y$ with respect to $u$ .

$$
\frac{dy}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}
$$

Differentiate $u = \tan 3x$ with respect to $x$ using the chain rule again.
Let $v = 3x$ . Then $u = \tan v$ . $\frac{du}{dv} = \sec^2 v = \sec^2 3x$ $\frac{dv}{dx} = 3$ $\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = \sec^2 3x \cdot 3 = 3\sec^2 3x$
Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .
$$ \frac{dy}{dx} = \frac{1}{2\sqrt{\tan 3x}} \cdot (3\sec^2 3x) = \frac{3\sec^2 3x}{2\sqrt{\tan 3x}} $$ The final answer is $\boxed{\frac{3\sec^2 3x}{2\sqrt{\tan 3x}}}$.
2. Find $\frac{dy}{dx}$ by using the product rule
The product rule states that if $y = u(x)v(x)$ , then $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$ .
- a) $y = 3^{4x} x^7$
Identify $u(x)$ and $v(x)$ .
Let $u = 3^{4x}$ and $v = x^7$.

Find $u'(x)$ and $v'(x)$ .

$$
u' = \frac{d}{dx}(3^{4x}) = 3^{4x} \ln 3 \cdot 4 = 4 \cdot 3^{4x} \ln 3
$$
$$
v' = \frac{d}{dx}(x^7) = 7x^6
$$

Apply the product rule.

$$
\frac{dy}{dx} = (4 \cdot 3^{4x} \ln 3) x^7 + 3^{4x} (7x^6)
$$

Factor out common terms.

$$
\frac{dy}{dx} = 3^{4x} x^6 (4x \ln 3 + 7)
$$
The final answer is $\boxed{3^{4x} x^6 (4x \ln 3 + 7)}$.

b) $y = \ln x \cdot \sec x$

Identify $u(x)$ and $v(x)$ .
Let $u = \ln x$ and $v = \sec x$.

Find $u'(x)$ and $v'(x)$ .

$$
u' = \frac{d}{dx}(\ln x) = \frac{1}{x}
$$
$$
v' = \frac{d}{dx}(\sec x) = \sec x \tan x
$$

Apply the product rule.

$$
\frac{dy}{dx} = \left(\frac{1}{x}\right) \sec x + (\ln x) (\sec x \tan x)
$$

Factor out common terms.

$$
\frac{dy}{dx} = \sec x \left(\frac{1}{x} + \ln x \tan x\right)
$$
The final answer is $\boxed{\sec x \left(\frac{1}{x} + \ln x \tan x\right)}$.

c) $y = e^{3x} \cdot \log x$

Identify $u(x)$ and $v(x)$ .
Assume $\log x = \ln x$ (natural logarithm).
Let $u = e^{3x}$ and $v = \ln x$ .

Find $u'(x)$ and $v'(x)$ .

$$
u' = \frac{d}{dx}(e^{3x}) = e^{3x} \cdot 3 = 3e^{3x}
$$
$$
v' = \frac{d}{dx}(\ln x) = \frac{1}{x}
$$

Apply the product rule.

$$
\frac{dy}{dx} = (3e^{3x}) \ln x + e^{3x} \left(\frac{1}{x}\right)
$$

Factor out common terms.

$$
\frac{dy}{dx} = e^{3x} \left(3 \ln x + \frac{1}{x}\right)
$$
The final answer is $\boxed{e^{3x} \left(3 \ln x + \frac{1}{x}\right)}$.

d) $y = \frac{1}{4x^3} \cdot \tan x$

Identify $u(x)$ and $v(x)$ .
Rewrite $u(x)$ as $\frac{1}{4}x^{-3}$ .
Let $u = \frac{1}{4}x^{-3}$ and $v = \tan x$ .

Find $u'(x)$ and $v'(x)$ .

$$
u' = \frac{d}{dx}\left(\frac{1}{4}x^{-3}\right) = \frac{1}{4}(-3)x^{-4} = -\frac{3}{4x^4}
$$
$$
v' = \frac{d}{dx}(\tan x) = \sec^2 x
$$

Apply the product rule.

$$
\frac{dy}{dx} = \left(-\frac{3}{4x^4}\right) \tan x + \left(\frac{1}{4x^3}\right) \sec^2 x
$$

Combine terms with a common denominator.
$$ \frac{dy}{dx} = \frac{-3 \tan x}{4x^4} + \frac{x \sec^2 x}{4x^4} = \frac{x \sec^2 x - 3 \tan x}{4x^4} $$ The final answer is $\boxed{\frac{x \sec^2 x - 3 \tan x}{4x^4}}$.
3. Find $\frac{dy}{dx}$ by using the quotient rule
The quotient rule states that if $y = \frac{u(x)}{v(x)}$ , then $\frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ .
- a) $y = \frac{x^8}{\ln x}$
Identify $u(x)$ and $v(x)$ .
Let $u = x^8$ and $v = \ln x$.

Find $u'(x)$ and $v'(x)$ .

$$
u' = \frac{d}{dx}(x^8) = 8x^7
$$
$$
v' = \frac{d}{dx}(\ln x) = \frac{1}{x}
$$

Apply the quotient rule.

$$
\frac{dy}{dx} = \frac{(8x^7)(\ln x) - (x^8)\left(\frac{1}{x}\right)}{(\ln x)^2}
$$

Simplify the expression.

$$
\frac{dy}{dx} = \frac{8x^7 \ln x - x^7}{(\ln x)^2} = \frac{x^7(8 \ln x - 1)}{(\ln x)^2}
$$
The final answer is $\boxed{\frac{x^7(8 \ln x - 1)}{(\ln x)^2}}$.

b) $y = \frac{\sin x}{x^4}$

Identify $u(x)$ and $v(x)$ .
Let $u = \sin x$ and $v = x^4$.

Find $u'(x)$ and $v'(x)$ .

$$
u' = \frac{d}{dx}(\sin x) = \cos x
$$
$$
v' = \frac{d}{dx}(x^4) = 4x^3
$$

Apply the quotient rule.

$$
\frac{dy}{dx} = \frac{(\cos x)(x^4) - (\sin x)(4x^3)}{(x^4)^2}
$$

Simplify the expression.

$$
\frac{dy}{dx} = \frac{x^4 \cos x - 4x^3 \sin x}{x^8} = \frac{x^3(x \cos x - 4 \sin x)}{x^8} = \frac{x \cos x - 4 \sin x}{x^5}
$$
The final answer is $\boxed{\frac{x \cos x - 4 \sin x}{x^5}}$.

c) $y = \frac{3^{-2x}}{x^2}$

Identify $u(x)$ and $v(x)$ .
Let $u = 3^{-2x}$ and $v = x^2$.

Find $u'(x)$ and $v'(x)$ .

$$
u' = \frac{d}{dx}(3^{-2x}) = 3^{-2x} \ln 3 \cdot (-2) = -2 \cdot 3^{-2x} \ln 3
$$
$$
v' = \frac{d}{dx}(x^2) = 2x
$$

Apply the quotient rule.

$$
\frac{dy}{dx} = \frac{(-2 \cdot 3^{-2x} \ln 3)(x^2) - (3^{-2x})(2x)}{(x^2)^2}
$$

Simplify the expression.

$$
\frac{dy}{dx} = \frac{-2x^2 \cdot 3^{-2x} \ln 3 - 2x \cdot 3^{-2x}}{x^4} = \frac{2x \cdot 3^{-2x}(-x \ln 3 - 1)}{x^4}
$$
$$
\frac{dy}{dx} = \frac{2 \cdot 3^{-2x}(-x \ln 3 - 1)}{x^3} = \frac{-2 \cdot 3^{-2x}(x \ln 3 + 1)}{x^3}
$$
The final answer is $\boxed{\frac{-2 \cdot 3^{-2x}(x \ln 3 + 1)}{x^3}}$.

d) $y = \frac{\sec x}{x^7}$

Identify $u(x)$ and $v(x)$ .
Let $u = \sec x$ and $v = x^7$.

Find $u'(x)$ and $v'(x)$ .

$$
u' = \frac{d}{dx}(\sec x) = \sec x \tan x
$$
$$
v' = \frac{d}{dx}(x^7) = 7x^6
$$

Apply the quotient rule.

$$
\frac{dy}{dx} = \frac{(\sec x \tan x)(x^7) - (\sec x)(7x^6)}{(x^7)^2}
$$

Simplify the expression.

$$
\frac{dy}{dx} = \frac{x^7 \sec x \tan x - 7x^6 \sec x}{x^{14}} = \frac{x^6 \sec x (x \tan x - 7)}{x^{14}}
$$
$$
\frac{dy}{dx} = \frac{\sec x (x \tan x - 7)}{x^8}
$$
The final answer is $\boxed{\frac{\sec x (x \tan x - 7)}{x^8}}$.

Find dy/dx by using chain rule Find dy/dx by using produce rule Find dy/dx by using Quotient. rule.

Step-by-step solution

Rewrite yyy as (sec⁡x)1/2(\sec x)^{1/2}(secx)1/2.

Differentiate yyy with respect to uuu and uuu with respect to xxx.

Apply the chain rule dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​.

Rewrite yyy as (4x2−3x)1/2(4x^2 - 3x)^{1/2}(4x2−3x)1/2.

Differentiate yyy with respect to uuu and uuu with respect to xxx.

Apply the chain rule dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​.

Rewrite yyy as (cos⁡x)−3(\cos x)^{-3}(cosx)−3.

Differentiate yyy with respect to uuu and uuu with respect to xxx.

Apply the chain rule dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​.

Rewrite yyy as (tan⁡3x)1/2(\tan 3x)^{1/2}(tan3x)1/2.

Differentiate yyy with respect to uuu.

Differentiate u=tan⁡3xu = \tan 3xu=tan3x with respect to xxx using the chain rule again.

Apply the chain rule dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​.

Identify u(x)u(x)u(x) and v(x)v(x)v(x).

Find u′(x)u'(x)u′(x) and v′(x)v'(x)v′(x).

Apply the product rule.

Factor out common terms.

Identify u(x)u(x)u(x) and v(x)v(x)v(x).

Find u′(x)u'(x)u′(x) and v′(x)v'(x)v′(x).

Apply the product rule.

Factor out common terms.

Identify u(x)u(x)u(x) and v(x)v(x)v(x).

Find u′(x)u'(x)u′(x) and v′(x)v'(x)v′(x).

Apply the product rule.

Factor out common terms.

Identify u(x)u(x)u(x) and v(x)v(x)v(x).

Find u′(x)u'(x)u′(x) and v′(x)v'(x)v′(x).

Apply the product rule.

Combine terms with a common denominator.

Identify u(x)u(x)u(x) and v(x)v(x)v(x).

Find u′(x)u'(x)u′(x) and v′(x)v'(x)v′(x).

Apply the quotient rule.

Simplify the expression.

Identify u(x)u(x)u(x) and v(x)v(x)v(x).

Find u′(x)u'(x)u′(x) and v′(x)v'(x)v′(x).

Apply the quotient rule.

Simplify the expression.

Identify u(x)u(x)u(x) and v(x)v(x)v(x).

Find u′(x)u'(x)u′(x) and v′(x)v'(x)v′(x).

Apply the quotient rule.

Simplify the expression.

Identify u(x)u(x)u(x) and v(x)v(x)v(x).

Find u′(x)u'(x)u′(x) and v′(x)v'(x)v′(x).

Apply the quotient rule.

Simplify the expression.

Still stuck on this one?

More Mathematics Questions

Rewrite $y$ as $(\sec x)^{1/2}$ .

Differentiate $y$ with respect to $u$ and $u$ with respect to $x$ .

Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

Rewrite $y$ as $(4x^2 - 3x)^{1/2}$ .

Differentiate $y$ with respect to $u$ and $u$ with respect to $x$ .

Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

Rewrite $y$ as $(\cos x)^{-3}$ .

Differentiate $y$ with respect to $u$ and $u$ with respect to $x$ .

Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

Rewrite $y$ as $(\tan 3x)^{1/2}$ .

Differentiate $y$ with respect to $u$ .

Differentiate $u = \tan 3x$ with respect to $x$ using the chain rule again.

Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .

Identify $u(x)$ and $v(x)$ .

Find $u'(x)$ and $v'(x)$ .

Identify $u(x)$ and $v(x)$ .

Find $u'(x)$ and $v'(x)$ .

Identify $u(x)$ and $v(x)$ .

Find $u'(x)$ and $v'(x)$ .

Identify $u(x)$ and $v(x)$ .

Find $u'(x)$ and $v'(x)$ .

Identify $u(x)$ and $v(x)$ .

Find $u'(x)$ and $v'(x)$ .

Identify $u(x)$ and $v(x)$ .

Find $u'(x)$ and $v'(x)$ .

Identify $u(x)$ and $v(x)$ .

Find $u'(x)$ and $v'(x)$ .

Identify $u(x)$ and $v(x)$ .

Find $u'(x)$ and $v'(x)$ .