How do you differentiate #log_2(x^2sinx^2) #?

Question

How do you differentiate #log_2(x^2sinx^2)   #?

Samantha Adkison · Accepted Answer

#= 2/(ln 2) ( 1/x + x cot x^2)#

#(log_2(x^2sinx^2))'#

first we need to shift the base into natural logs as these are just made for calculus

so using #log_a b = (log_c b)/(log_c a)#

so #log_2(x^2sinx^2) = (ln (x^2sinx^2))/(ln 2) = 1/(ln 2)ln (x^2sinx^2)#

then we use another standard result namely that

#(ln f(x))' = 1/(f(x))*f'(x)#

so #( 1/(ln 2)ln (x^2sinx^2))' = 1/(ln 2)( ln (x^2sinx^2))' #

#= 1/(ln 2)1/ (x^2sinx^2) (x^2sinx^2)' \\qquad square#

for # (x^2sinx^2)'# we use the product rule

# (x^2sinx^2)' = (x^2)'sinx^2 + x^2 (sinx^2)' #

#= 2x sinx^2 + x^2 (sinx^2)' \qquad star#

for #(sinx^2)'# we use the chain rule so

#(sinx^2)' = cos x^2 (2x) = 2x cos x^2#

pop that back into #star # to get

# (x^2sinx^2)' = 2x sinx^2 + x^2 * 2x cos x^2 # #\implies (x^2sinx^2)' = 2x sinx^2 + 2x^3 cos x^2 #

pop that back into #square# for

#( 1/(ln 2)ln (x^2sinx^2))' = 1/(ln 2)1/ (x^2sinx^2) (x^2sinx^2)'#

#= 1/(ln 2)1/ (x^2sinx^2) (2x sinx^2 + 2x^3 cos x^2)#

this can be simplified in any number of ways, so i'm going for brevity

#= 2/(ln 2) ( 1/x + x cot x^2)#

Scarlett Allison · Answer

undefined To differentiate $ \log_2(x^2\sin(x^2)) $, use the chain rule and the derivative of the natural logarithm function:

$$ \frac{d}{dx}(\log_2(u)) = \frac{1}{u \ln(2)} \cdot \frac{du}{dx} $$

Let $ u = x^2\sin(x^2) $, then differentiate $ u $ with respect to $ x $:

$$ \frac{du}{dx} = \frac{d}{dx}(x^2\sin(x^2)) $$

Using the product rule and the chain rule:

$$ \frac{du}{dx} = 2x\sin(x^2) + x^2\cos(x^2) \cdot 2x $$

$$ \frac{du}{dx} = 2x\sin(x^2) + 2x^3\cos(x^2) $$

Now, substitute $ u $ and $ \frac{du}{dx} $ back into the formula for differentiating $ \log_2(u) $:

$$ \frac{d}{dx}(\log_2(x^2\sin(x^2))) = \frac{1}{x^2\sin(x^2)\ln(2)} \cdot \left(2x\sin(x^2) + 2x^3\cos(x^2)\right) $$

$$ \frac{d}{dx}(\log_2(x^2\sin(x^2))) = \frac{2x\sin(x^2) + 2x^3\cos(x^2)}{x^2\sin(x^2)\ln(2)} $$

Christopher Agresta · Answer

undefined To differentiate $ \log_2(x^2 \sin(x^2)) $, we can use the chain rule and the properties of logarithms. The derivative of $ \log_b(u) $ with respect to $ x $ is $ \frac{1}{u} \cdot \frac{du}{dx} $.

Let $ u = x^2 \sin(x^2) $. Then, applying the chain rule:

$$ \frac{d}{dx}(\log_2(u)) = \frac{1}{u} \cdot \frac{du}{dx} $$

Now, differentiate $ u = x^2 \sin(x^2) $ with respect to $ x $:

$$ \frac{du}{dx} = \frac{d}{dx}(x^2 \sin(x^2)) $$

Using the product rule:

$$ \frac{du}{dx} = 2x \sin(x^2) + x^2 \cdot \frac{d}{dx}(\sin(x^2)) $$

$$ \frac{du}{dx} = 2x \sin(x^2) + x^2 \cdot \cos(x^2) \cdot 2x $$

$$ \frac{du}{dx} = 2x \sin(x^2) + 2x^3 \cos(x^2) $$

Now, substitute $ u $ and $ \frac{du}{dx} $ into the expression for the derivative:

$$ \frac{d}{dx}(\log_2(x^2 \sin(x^2))) = \frac{1}{x^2 \sin(x^2)} \cdot (2x \sin(x^2) + 2x^3 \cos(x^2)) $$

$$ \frac{d}{dx}(\log_2(x^2 \sin(x^2))) = \frac{2x \sin(x^2) + 2x^3 \cos(x^2)}{x^2 \sin(x^2)} $$

So, the derivative of $ \log_2(x^2 \sin(x^2)) $ with respect to $ x $ is $ \frac{2x \sin(x^2) + 2x^3 \cos(x^2)}{x^2 \sin(x^2)} $.