Question

How do you differentiate `y = 2^xlog_2(x^4)`?

Answer 1

`f = 2^x, g = log_2(x^4)->f'=2^xln2, g' = 1/(x^4ln2)*4x^3`
`y'=fg'+gf'=(2^x4x^3)/(x^4ln2) +( 2^xln2)(log_2x^4)`

Explanation:

Use the product rule to differentiate y. Call the first one be f and the second one be g and take the derivatives of each of them then put it in to the product rule

Answer 2

`\frac{d}{dx}(2^x\log _2(x^4))=\frac{2^x(x\log _2(x^4)\ln ^2(2)+4)}{x\ln (2)}`

Explanation:

`\frac{d}{dx}(2^x\log _2(x^4))`

Applying product rule as: `(f\cdot g)^'=f^'\cdot g+f\cdot g^'`
`f=2^x,g=\log _2(x^4)`

`=\frac{d}{dx}(2^x)\log _2(x^4)+\frac{d}{dx}(\log _2(x^4))2^x`

`\frac{d}{dx}(2^x)=2^x\ln(2)`

`\frac{d}{dx}(\log _2(x^4))=\frac{4}{x\ln (2)}`

finally,
`=2^x\ln(2)\log _2(x^4)+\frac{4}{x\ln (2)}2^x`

Simplifying,
`=\frac{2^x(x\log _2(x^4)\ln ^2(2)+4)}{x\ln (2)}`

Answer 3

To solve this problem it would help if we remember these two intermediate results (proof given below),
Result 1: `\frac{d}{dx}a^x=a^x\ln(a)`
Result 2: `\frac{d}{dx}\log_b(x^n)=\frac{n}{x\ln(b)}`

`y = 2^x\log_2(x^4);`

Applying the Chain Rule,
`\frac{dy}{dx} = [\frac{d}{dx}(2^x)]\log_2(x^4) + 2^x[\frac{d}{dx}\log_2(x^4)]`

Apply the two results mentioned above,
`\frac{dy}{dx} = [2^x\ln(2)]log_2(x^4) + 2^x[\frac{4}{x\ln(2)}]`
`\qquad \quad = \ln(2)y+\frac{4}{\ln(2)}\frac{2^x}{x}`

Proof of Result 1:
Let `\quad y=a^x \qquad => \ln(y) = x\ln(a)`.
Differentiating once,
`1/y\frac{dy}{dx} = \ln(a) \qquad => \frac{dy}{dx} = y\ln(a)=a^x\ln(a)`

Proof of Result 2:
Let `\quad y=\log_b(x^n)`
Using the logarithm-base rule, `\log_b(x^n) = \log_e(x^n)/\log_e(b)`, we can write `y` as -

`y = \frac{\ln(x^n)}{\ln(b)}=\frac{n}{\ln(b)}\ln(x)`

`\frac{dy}{dx} = \frac{n}{\ln(b)}1/x=\frac{n}{x\ln(b)}`