List of Derivative Problems

The function does not depend on $x$: for every $x$ its value is the same number $3a.$ The derivative of a constant is zero.

${y'}={(3a)'}={0}$

Answer: ${y'=0}$

Differentiate term by term. The constant factor $5$ stays, ${(x)'=1},$ and the derivative of the number $4$ is zero.

${y'}={(5x)'-(4)'}={5\cdot 1-0}={5}$

Answer: ${y'=5}$

No quotient rule is needed, because the denominator is a number. Divide each term by $6$:

${y}={\frac{\sqrt2}{6}\,x-\frac12}$

The number $\frac{\sqrt2}{6}$ is a constant factor and $\frac12$ is a constant:

${y'}={\frac{\sqrt2}{6}\cdot 1-0}={\frac{\sqrt2}{6}}$

Answer: ${y'=\frac{\sqrt2}{6}}$

Write ${\frac{2x}{5}=\frac25x}$ and differentiate term by term:

${y'}={(8)'-\left(\frac25x\right)'}={0-\frac25\cdot 1}={-\frac25}$

Answer: ${y'=-\frac25}$

By the power rule ${(x^2)'=2x}.$ The constant factor $0.5$ stays:

${y'}={0.5\cdot 2x}={x}$

Answer: ${y'=x}$

The number $\sqrt7$ is a constant factor in front of $x,$ just like $3$ in front of $x^2.$

${y'}={3\cdot 2x+\sqrt7\cdot 1+0}={6x+\sqrt7}$

Answer: ${y'=6x+\sqrt7}$

Differentiate the polynomial term by term with ${(x^n)'=nx^{n-1}}:$

${y'}={0-2x+1-3\cdot 4x^3}={-12x^3-2x+1}$

At the end we put the terms in order of decreasing powers of $x.$

Answer: ${y'=-12x^3-2x+1}$

${y'}={-3x^2+4\cdot 2x-0}={-3x^2+8x}$

Answer: ${y'=-3x^2+8x}$

${y'}={5\cdot 3x^2-\sqrt2\cdot 2x+6}={15x^2-2\sqrt2\,x+6}$

Answer: ${y'=15x^2-2\sqrt2\,x+6}$

The exponents $n$ and ${3-n}$ are numbers that do not depend on $x,$ so the power rule works here too: the exponent comes down as a factor and the new exponent is one less.

${y'}={2\cdot nx^{n-1}}+{(3-n)x^{3-n-1}+0}={2nx^{n-1}+(3-n)x^{2-n}}$

Do not forget the factor $n$ in the first term: ${(2x^n)'}$ is ${2nx^{n-1}},$ not ${2x^{n-1}}.$

Answer: ${y'}={2nx^{n-1}+(3-n)x^{2-n}}$

This is a product, so we use ${(f\cdot g)'=f'g+fg'}$ with ${f=x-1}$ and ${g=x^2+3}:$

${y'}={1\cdot(x^2+3)}+{(x-1)\cdot 2x}={x^2+3+2x^2-2x}={3x^2-2x+3}$

Check another way: ${y=x^3-x^2+3x-3},$ so ${y'=3x^2-2x+3}$, the same result.

Answer: ${y'=3x^2-2x+3}$

The power rule also works for a negative exponent: ${(x^{-8})'=-8x^{-9}}.$

${y'}={\frac{4x^3}{4}-(-8)x^{-9}}={x^3+8x^{-9}}={x^3+\frac{8}{x^9}}$

Watch the signs: the minus in front of $x^{-8}$ times the minus of the exponent gives a plus.

Answer: ${y'=x^3+\frac{8}{x^9}}$

Write both fractions as powers: ${y=\frac12x+2x^{-1}}.$ Then

${y'}={\frac12\cdot 1+2\cdot(-1)x^{-2}}={\frac12-\frac{2}{x^2}}$

Answer: ${y'=\frac12-\frac{2}{x^2}}$

Expand the bracket with the square of a sum formula:

${y}={x^2+4x+4-5x^3}$

${y'}={2x+4-15x^2}={-15x^2+2x+4}$

You can also skip the expansion and use the chain rule: ${\left((x+2)^2\right)'}={2(x+2)\cdot 1}={{2x+4}.}$

Answer: ${y'=-15x^2+2x+4}$

With three factors it is easier to multiply them out first and then differentiate the polynomial:

${y}={(x^3-3x)(x^3+1)}={x^6-3x^4+x^3-3x}$

${y'}={6x^5-12x^3+3x^2-3}$

Answer: ${y'=6x^5-12x^3+3x^2-3}$

The letter $a$ is a number, so we treat it like one. Expand the bracket:

${y}={a^2x^2+2ax+1-2ax^3}$

${y'}={a^2\cdot 2x+2a\cdot 1}+{0-2a\cdot 3x^2}={-6ax^2+2a^2x+2a}$

The term ${-2ax^3}$ has a minus sign, so ${-6ax^2}$ in the answer has a minus too.

Answer: ${y'=-6ax^2+2a^2x+2a}$

Write ${\sqrt{x}=x^{\frac12}}$ and multiply:

${y}={2x^{\frac52}-4x^{\frac32}}$

${y'}={2\cdot\frac52x^{\frac32}-4\cdot\frac32x^{\frac12}}={5x^{\frac32}-6x^{\frac12}}={(5x-6)\sqrt{x}}$

In the last step we took ${x^{\frac12}=\sqrt{x}}$ out of the bracket, because ${x^{\frac32}=x\sqrt{x}}.$

Answer: ${y'=(5x-6)\sqrt{x}}$

Use the quotient rule ${\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}}$ with ${f=\sqrt{x}},$ ${f'=\frac{1}{2\sqrt{x}}}$ and ${g=x+3},$ ${g'=1}:$

${y'}={\frac{\frac{1}{2\sqrt{x}}(x+3)-\sqrt{x}\cdot 1}{(x+3)^2}}$

Multiply the numerator and the denominator by $2\sqrt{x}$ to clear the fraction in the numerator. Since ${\sqrt{x}\cdot 2\sqrt{x}=2x},$

${y'}={\frac{x+3-2x}{2\sqrt{x}\,(x+3)^2}}={\frac{3-x}{2\sqrt{x}\,(x+3)^2}}$

Answer: ${y'=\frac{3-x}{2\sqrt{x}\,(x+3)^2}}$

${y'}={8}$

${y''}={(8)'}={0}$

Answer: ${y''=0}$

${y'}={24x-16}$

${y''}={(24x-16)'}={24}$

Answer: ${y''=24}$

Write ${\frac1x=x^{-1}}$ and differentiate twice:

${y'}={3x^2-8x-x^{-2}}={3x^2-8x-\frac{1}{x^2}}$

${y''}={6x-8-(-2)x^{-3}}={6x-8+\frac{2}{x^3}}$

Answer: ${y''=6x-8+\frac{2}{x^3}}$

Find the first derivative with the quotient rule:

${y'}={\frac{1\cdot(x+5)-x\cdot 1}{(x+5)^2}}={\frac{5}{(x+5)^2}}={5(x+5)^{-2}}$

The second time we do not need the quotient rule: we wrote $y'$ as a power and differentiate it with the chain rule (the inner function ${x+5}$ has derivative $1$):

${y''}={5\cdot(-2)(x+5)^{-3}\cdot 1}={-\frac{10}{(x+5)^3}}$

Answer: ${y''=-\frac{10}{(x+5)^3}}$

This is a composite function: the outer one is $\cos,$ the inner one is ${3x}$ with derivative $3.$ So

${y'}={-\sin 3x\cdot 3}={-3\sin 3x}$

${y''}={-3\cos 3x\cdot 3}={-9\cos 3x}$

Answer: ${y''=-9\cos 3x}$

The notation $\sin^3x$ means ${(\sin x)^3}.$ The outer function is the cube, the inner one is ${\sin x}:$

${y'}={3\sin^2x\cdot\cos x}$

Now $y'$ is the product of ${3\sin^2x}$ and ${\cos x}.$ By the product rule, with ${(\sin^2x)'=2\sin x\cos x}:$

${y''}={3\cdot 2\sin x\cos x\cdot\cos x}+{3\sin^2x\cdot(-\sin x)}={6\sin x\cos^2x-3\sin^3x}$

Substitute ${\cos^2x=1-\sin^2x}$ so that only sine is left:

${y''}={6\sin x(1-\sin^2x)-3\sin^3x}={6\sin x-9\sin^3x}$

Answer: ${y''=6\sin x-9\sin^3x}$

Write the function as a power: ${y=\frac14(x+4)^{\frac12}}.$ The inner function ${x+4}$ has derivative $1:$

${y'}={\frac14\cdot\frac12(x+4)^{-\frac12}}={\frac18(x+4)^{-\frac12}}={\frac{1}{8\sqrt{x+4}}}$

${y''}={\frac18\cdot\left(-\frac12\right)(x+4)^{-\frac32}}={-\frac{1}{16\sqrt{(x+4)^3}}}$

That is why we take ${x\gt -4}:$ at ${x=-4}$ the function is defined, but its derivatives are not, because the denominator becomes zero.

Answer: ${y''=-\frac{1}{16\sqrt{(x+4)^3}}}$