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

3 Answers
Mar 1, 2016

Answer:

#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

Mar 1, 2016

Answer:

#\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)}#

Mar 1, 2016

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)}#