Use \(\dfrac{d}{dx}(ax)=a\).

So \(f\,\!'(x)=-4\).

Use \(\dfrac{d}{dx}(ax)=a\).

So \(f\,\!'(x)=-15\).

Use \(\dfrac{d}{dx}(ax)=a\).

So \(f\,\!'(x)=-\dfrac12\).

Use \(\dfrac{d}{dx}(ax)=a\).

So \(\dfrac{dy}{dx}=\dfrac{2}{9}\).

Use \(\dfrac{d}{dx}(ax)=a\).

So \(\dfrac{dy}{dx}=-\dfrac23\).

Use \(\dfrac{d}{dx}(x)=1\).

So \(\dfrac{d}{dx}(-x)=-1\).

Use the power rule \(\dfrac{d}{dx}(x^n)=nx^{n-1}\).

So \(f\,\!'(x)=3x^2\).

Use the power rule.

\(\dfrac{d}{dx}(4x^2)=4\cdot2x=8x\).

Use the power rule.

\(f\,\!'(x)=9x^8\).

Use the power rule.

\(\dfrac{dy}{dx}=12\cdot7x^6=84x^6\).

Use the power rule.

\(f\,\!'(x)=\dfrac45\cdot10x^9=8x^9\).

Use the power rule.

\(\dfrac{dy}{dx}=\dfrac57\cdot21x^{20}=15x^{20}\).

Write \(\sqrt{x}=x^{1/2}\).

Then \(f\,\!'(x)=\dfrac12 x^{-1/2}=\dfrac{1}{2\sqrt{x}}\).

Use the power rule.

\(\dfrac{dy}{dx}=2\cdot\dfrac12 x^{-1/2}=x^{-1/2}=\dfrac{1}{\sqrt{x}}\).

Write \(-4\sqrt{x}=-4x^{1/2}\).

Then \(f\,\!'(x)=-4\cdot\dfrac12 x^{-1/2}=-2x^{-1/2}=\dfrac{-2}{\sqrt{x}}\).

Write \(5\sqrt{x}=5x^{1/2}\).

Then \(\dfrac{dy}{dx}=5\cdot\dfrac12 x^{-1/2}=\dfrac{5}{2\sqrt{x}}\).

Use the power rule:

\(\dfrac{d}{dx}(x^{1/3})=\dfrac13 x^{-2/3}\).

Use the power rule:

\(\dfrac{d}{dx}(4x^{-1/4})=4\cdot\left(-\dfrac14\right)x^{-5/4}=-x^{-5/4}\).

Rewrite \(\dfrac{2}{x}=2x^{-1}\).

Then \(f\,\!'(x)=2(-1)x^{-2}=-\dfrac{2}{x^2}\).

Rewrite \(\dfrac{7}{x^2}=7x^{-2}\).

Then \(f\,\!'(x)=7(-2)x^{-3}=-\dfrac{14}{x^3}\).

Use the power rule:

\(f\,\!'(x)=3(-1)x^{-2}=-\dfrac{3}{x^2}\).

Rewrite \(-\dfrac{5}{x}=-5x^{-1}\).

Then \(\dfrac{dy}{dx}=-5(-1)x^{-2}=\dfrac{5}{x^2}\).

Rewrite \(\dfrac{1}{x^2}=x^{-2}\).

Then \(\dfrac{d}{dx}(x^{-2})=-2x^{-3}=-\dfrac{2}{x^3}\).

Rewrite \(\dfrac{8}{x^3}=8x^{-3}\).

Then \(\dfrac{dy}{dx}=8(-3)x^{-4}=-\dfrac{24}{x^4}\).

Rewrite \(\dfrac{3}{\sqrt{x}}=3x^{-1/2}\).

Then \(\dfrac{dy}{dx}=3\left(-\dfrac12\right)x^{-3/2}=-\dfrac{3}{2x^{3/2}}\).

Rewrite \(\dfrac{1}{\sqrt[3]{x}}=x^{-1/3}\).

Then \(\dfrac{dy}{dx}=-\dfrac13 x^{-4/3}\).

Rewrite \(\dfrac{9}{\sqrt[3]{x}}=9x^{-1/3}\).

Then \(\dfrac{dy}{dx}=9\left(-\dfrac13\right)x^{-4/3}=-3x^{-4/3}\).

Rewrite \(-\dfrac{4}{\sqrt{x}}=-4x^{-1/2}\).

Then \(\dfrac{dy}{dx}=-4\left(-\dfrac12\right)x^{-3/2}=2x^{-3/2}=\dfrac{2}{x^{3/2}}\).

Use the power rule.

\(f\,\!'(x)=7(-4)x^{-5}=-28x^{-5}=\dfrac{-28}{x^5}\).

Simplify first:

\(x\times x^5=x^6\).

Then differentiate:

\(\dfrac{dy}{dx}=6x^5\).

Simplify first:

\(\dfrac{x^9}{x^3}=x^6\).

Then differentiate:

\(\dfrac{d}{dx}(x^6)=6x^5\).

Simplify first:

\(x^7\times x^4=x^{11}\).

Then differentiate:

\(\dfrac{d}{dx}(x^{11})=11x^{10}\).

Simplify first:

\(\dfrac{8x^{12}}{2x^4}=4x^8\).

Then differentiate:

\(\dfrac{d}{dx}(4x^8)=32x^7\).

Simplify first:

\(12x^8\div 3x=4x^7\).

Then differentiate:

\(\dfrac{dy}{dx}=28x^6\).

Simplify first:

\(4x\times5x^2=20x^3\).

Then differentiate:

\(\dfrac{dy}{dx}=60x^2\).

Rewrite first:

\(\sqrt[3]{8x^2}=(8x^2)^{1/3}=2x^{2/3}\).

Then differentiate:

\(\dfrac{d}{dx}(2x^{2/3})=2\cdot\dfrac23 x^{-1/3}=\dfrac{4}{3}x^{-1/3}\).

Simplify first:

\(\dfrac{x}{x^6}=x^{-5}\).

Then differentiate:

\(\dfrac{d}{dx}(x^{-5})=-5x^{-6}\).

Simplify first:

\(5x^9\times2x=10x^{10}\).

Then differentiate:

\(\dfrac{dy}{dx}=100x^9\).

Simplify first:

\(x^{3/2}\div x^{1/2}=x\).

Then differentiate:

\(\dfrac{dy}{dx}=1\).

Simplify first:

\((-x^2)(-3x^3)=3x^5\).

Then differentiate:

\(\dfrac{dy}{dx}=15x^4\).

Rewrite \(-\dfrac{1}{2x^3}= -\dfrac12 x^{-3}\).

Then differentiate:

\(\dfrac{d}{dx}\left(-\dfrac12 x^{-3}\right)= -\dfrac12(-3)x^{-4}=\dfrac32 x^{-4}=\dfrac{3}{2x^4}\).

Use the power rule:

\(\dfrac{dy}{dx}=7(-4)x^{-5}=-28x^{-5}\).

Write \(16\sqrt{x}=16x^{1/2}\).

Then \(\dfrac{dy}{dx}=16\cdot\dfrac12 x^{-1/2}=8x^{-1/2}=\dfrac{8}{\sqrt{x}}\).

Write \(y=\dfrac13 x^{1/2}\).

Then \(\dfrac{dy}{dx}=\dfrac13\cdot\dfrac12 x^{-1/2}=\dfrac{1}{6\sqrt{x}}\).

Simplify first:

\(\dfrac{2x^5+7x}{x}=2x^4+7\).

Then differentiate:

\(f\,\!'(x)=8x^3\).

Simplify first:

\(\dfrac{9x^3-3x}{x}=9x^2-3\).

Then differentiate:

\(f\,\!'(x)=18x\).

Simplify first:

\(\dfrac{24x^3-8x}{x}=24x^2-8\).

Then differentiate:

\(f\,\!'(x)=48x\).

Simplify first:

\((-x^2)(-3x^3)=3x^5\).

Then differentiate:

\(\dfrac{dy}{dx}=15x^4\).

First expand:

\(y=x^2+3x\).

Now differentiate term by term:

\(\dfrac{dy}{dx}=2x+3\).

First expand:

\(y=x^2-3x-10\).

Now differentiate:

\(\dfrac{dy}{dx}=2x-3\).

First expand:

\(y=2x^2+7x-4\).

Now differentiate:

\(\dfrac{dy}{dx}=4x+7\).

First expand:

\(y=3x^3-5x\).

Now differentiate:

\(\dfrac{dy}{dx}=9x^2-5\).

First expand:

\(y=2x^3+2x\).

Now differentiate:

\(\dfrac{dy}{dx}=6x^2+2\).

Write \(\sqrt{x}=x^{1/2}\).

Then \(y=x^{1/2}\times x^2=x^{5/2}\).

Now differentiate:

\(\dfrac{dy}{dx}=\dfrac{5}{2}x^{3/2}\).

Write \(\sqrt{x}=x^{1/2}\).

Then \(y=3x^{3/2}\).

Now differentiate:

\(\dfrac{dy}{dx}=3\cdot \dfrac{3}{2}x^{1/2}=\dfrac{9}{2}x^{1/2}\).

Write \(\dfrac{2}{\sqrt{x}}=2x^{-1/2}\).

Then \(y=2x^{-1/2}\times x^3=2x^{5/2}\).

Now differentiate:

\(\dfrac{dy}{dx}=2\cdot \dfrac{5}{2}x^{3/2}=5x^{3/2}\).

Simplify first:

\(y=x^2\).

Now differentiate:

\(\dfrac{dy}{dx}=2x\).

Simplify first:

\(y=2x^4-x\).

Now differentiate:

\(\dfrac{dy}{dx}=8x^3-1\).

Simplify first:

\(y=x^2+2\).

Now differentiate:

\(\dfrac{dy}{dx}=2x\).

Simplify first:

\(y=2x-4\).

Now differentiate:

\(\dfrac{dy}{dx}=2\).

Simplify first:

\(y=10x^5\).

Now differentiate:

\(\dfrac{dy}{dx}=50x^4\).

Expand first:

\(y=\sqrt{x}\cdot \sqrt{x}+4\sqrt{x}=x+4x^{1/2}\).

Now differentiate:

\(\dfrac{dy}{dx}=1+4\cdot \dfrac12 x^{-1/2}=1+\dfrac{2}{\sqrt{x}}\).

Use the difference of squares:

\((x^2-1)(x^2+1)=x^4-1\).

Now differentiate:

\(\dfrac{dy}{dx}=4x^3\).

Simplify first:

\(y=4x^4\).

Now differentiate:

\(\dfrac{dy}{dx}=16x^3\).

Write \(\sqrt{x}=x^{1/2}\).

Then

\(y=\dfrac{x^2}{x^{1/2}}+\dfrac{3x}{x^{1/2}}=x^{3/2}+3x^{1/2}\).

Now differentiate:

\(\dfrac{dy}{dx}=\dfrac{3}{2}x^{1/2}+\dfrac{3}{2}x^{-1/2}\).

Use the difference of squares:

\((2x-1)(2x+1)=4x^2-1\).

Now differentiate:

\(\dfrac{dy}{dx}=8x\).

Write \(\dfrac{3}{x^2}=3x^{-2}\).

Then \(y=3x^{-2}\cdot x^5=3x^3\).

Now differentiate:

\(\dfrac{dy}{dx}=9x^2\).

Simplify first:

\(y=x^2-1\).

Now differentiate:

\(\dfrac{dy}{dx}=2x\).

Use the chain rule:

\(\dfrac{d}{dx}(u^2)=2u\cdot \dfrac{du}{dx}\).

Here \(u=2x+3\), so \(\dfrac{du}{dx}=2\).

Therefore

\(\dfrac{dy}{dx}=2(2x+3)\cdot 2=8x+12\).

Use the chain rule:

\(\dfrac{d}{dx}(u^3)=3u^2\cdot \dfrac{du}{dx}\).

Here \(u=3x-1\), so \(\dfrac{du}{dx}=3\).

Therefore

\(\dfrac{dy}{dx}=3(3x-1)^2\cdot 3=9(3x-1)^2\).

Let \(u=x^2+4\).

Then \(y=u^5\), so \(\dfrac{dy}{dx}=5u^4\cdot \dfrac{du}{dx}\).

Since \(\dfrac{du}{dx}=2x\),

\(\dfrac{dy}{dx}=10x(x^2+4)^4\).

Write \(y=(5x-2)^{1/2}\).

Using the chain rule,

\(\dfrac{dy}{dx}=\dfrac12(5x-2)^{-1/2}\cdot 5=\dfrac{5}{2\sqrt{5x-2}}\).

Write \(y=(2x+1)^{-1}\).

Use the chain rule:

\(\dfrac{dy}{dx}=-1(2x+1)^{-2}\cdot 2=-\dfrac{2}{(2x+1)^2}\).

Let \(u=4-3x\).

Then \(\dfrac{du}{dx}=-3\).

So

\(\dfrac{dy}{dx}=4(4-3x)^3(-3)=-12(4-3x)^3\).

Let \(u=x^3+2\).

Then \(\dfrac{du}{dx}=3x^2\).

So

\(\dfrac{dy}{dx}=2(x^3+2)(3x^2)=6x^2(x^3+2)\).

Let \(u=2x^2-1\).

Then \(\dfrac{du}{dx}=4x\).

So

\(\dfrac{dy}{dx}=3(2x^2-1)^2(4x)=12x(2x^2-1)^2\).

Let \(u=7-x\).

Then \(\dfrac{du}{dx}=-1\).

So

\(\dfrac{dy}{dx}=5(7-x)^4(-1)=-5(7-x)^4\).

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

Using the chain rule,

\(\dfrac{dy}{dx}=\dfrac12(x^2+1)^{-1/2}\cdot 2x=\dfrac{x}{\sqrt{x^2+1}}\).

Write \(y=(3x+4)^{-1/2}\).

Using the chain rule,

\(\dfrac{dy}{dx}=-\dfrac12(3x+4)^{-3/2}\cdot 3=-\dfrac{3}{2(3x+4)^{3/2}}\).

Let \(u=x^2+x\).

Then \(\dfrac{du}{dx}=2x+1\).

So

\(\dfrac{dy}{dx}=4(x^2+x)^3(2x+1)\).

Let \(u=5x-2\).

Then \(\dfrac{du}{dx}=5\).

So

\(\dfrac{dy}{dx}=6(5x-2)^5\cdot 5=30(5x-2)^5\).

Let \(u=1-4x^2\).

Then \(\dfrac{du}{dx}=-8x\).

So

\(\dfrac{dy}{dx}=3(1-4x^2)^2(-8x)=-24x(1-4x^2)^2\).

Write \(y=(2-x^3)^{1/2}\).

Then

\(\dfrac{dy}{dx}=\dfrac12(2-x^3)^{-1/2}\cdot(-3x^2)=-\dfrac{3x^2}{2\sqrt{2-x^3}}\).

Let \(u=x^4-1\).

Then \(\dfrac{du}{dx}=4x^3\).

So

\(\dfrac{dy}{dx}=2(x^4-1)(4x^3)=8x^3(x^4-1)\).

Write \(y=(x^2+3)^{-1}\).

Then

\(\dfrac{dy}{dx}=-1(x^2+3)^{-2}(2x)=-\dfrac{2x}{(x^2+3)^2}\).

Use the chain rule:

\(\dfrac{dy}{dx}=\dfrac13(2x+1)^{-2/3}\cdot 2=\dfrac{2}{3}(2x+1)^{-2/3}\).

Let \(u=x^2-5x+1\).

Then \(\dfrac{du}{dx}=2x-5\).

So

\(\dfrac{dy}{dx}=2(x^2-5x+1)(2x-5)\).

Let \(u=3-2x\).

Then \(\dfrac{du}{dx}=-2\).

So

\(\dfrac{dy}{dx}=-2(3-2x)^{-3}(-2)=4(3-2x)^{-3}=\dfrac{4}{(3-2x)^3}\).