Question

What is the derivative of $ {{2}^{{{\left( \sin \,x \right)}^{2}}}} $ with respect to $ \sin \,x $ ?
A. $ \sin \,x\,{{2}^{{{\left( \sin x \right)}^{2}}}}\ln 4 $
B. $ 2\sin \,x\,{{2}^{{{\left( \sin x \right)}^{2}}}}\ln 2 $
C. $ \ln \left( \sin \,x \right)\,{{2}^{{{\left( \sin x \right)}^{2}}}} $
D. $ 2\sin \,x\,\cos x\,{{2}^{{{\left( \sin x \right)}^{2}}}} $

Answer 1

Hint: Here we have been given a value whose derivative is to be found with respect to $ \sin \,x $ . Firstly we will let both the values equal to some variable then we will differentiate each of them with respect to $ x $ using the formula of them. Finally we will divide both the values obtained and get our desired answer.

Complete step-by-step answer:
We have to find the derivative of the value given below:
 $ {{2}^{{{\left( \sin \,x \right)}^{2}}}} $
We have to differentiate the above value with respect to $ \sin x $
Let us assume them as follows:
 $ u={{2}^{{{\left( \sin x \right)}^{2}}}} $ …… $ \left( 1 \right) $
 $ v=\sin x $ …… $ \left( 2 \right) $
Now let us differentiate equation (1) with respect to $ x $ as follows:
 $ u={{2}^{{{\left( \sin x \right)}^{2}}}} $
Take $ \log $ both sides
 $ \log u=\log {{2}^{{{\left( \sin x \right)}^{2}}}} $
As we know property of logarithm $ \log {{\left( b \right)}^{a}}=a\log \left( b \right) $ we will use it above,
 $ \log u={{\left( \sin x \right)}^{2}}\log 2 $ …. $ \left( 3 \right) $
As we know the differentiation of logarithm and $ \sin x $ is done as follows:
 $ \dfrac{d\left( logx \right)}{dx}=\dfrac{1}{x}\dfrac{d\left( x \right)}{dx} $ and $ \dfrac{d\left( sinx \right)}{dx}=\cos x $
Use above formula in equation (3) we get,
 $ \dfrac{d\left( \log u \right)}{dx}=\dfrac{d\left( \log 2\times {{\left( \sin x \right)}^{2}} \right)}{dx} $
 $ \Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\log 2\left( 2\sin \dfrac{d\left( cosx \right)}{dx} \right) $
So we get the value as
 $ \Rightarrow \dfrac{du}{dx}=u\times \log 2\times 2\sin x\times \cos x $
Now replace value from equation (1) above we get,
 $ \Rightarrow \dfrac{du}{dx}={{2}^{{{\left( \sin x \right)}^{2}}}}\times \log 2\times 2\sin x\cos x $ ….. $ \left( 4 \right) $
Next we will differentiate equation (2) with respect to $ x $ as follows:
 $ \dfrac{dv}{dx}=\cos x $ …. $ \left( 5 \right) $
Final step will be to divide equation (4) by equation (5) as below:
 $ \dfrac{\dfrac{du}{dx}}{\dfrac{dv}{dx}}=\dfrac{{{2}^{{{\left( \sin x \right)}^{2}}}}\times \log 2\times 2\sin x\cos x}{\cos x} $
 $ \Rightarrow \dfrac{du}{dv}={{2}^{{{\left( \sin x \right)}^{2}}}}\times \log 2\times 2\sin x $
Put the value from equation (1) and (2) above we get,
 $ \Rightarrow \dfrac{d\left( {{2}^{{{\left( \sin x \right)}^{2}}}} \right)}{d\left( \sin x \right)}=\sin x{{2}^{{{\left( \sin x \right)}^{2}}}}2\ln 2 $
Using logarithm property $ \log {{\left( b \right)}^{a}}=a\log \left( b \right) $ we can rewrite the above value as
  $ \Rightarrow \dfrac{d\left( {{2}^{{{\left( \sin x \right)}^{2}}}} \right)}{d\left( \sin x \right)}=\sin x{{2}^{{{\left( \sin x \right)}^{2}}}}\ln {{2}^{2}} $
 $ \Rightarrow \dfrac{d\left( {{2}^{{{\left( \sin x \right)}^{2}}}} \right)}{d\left( \sin x \right)}=\sin x{{2}^{{{\left( \sin x \right)}^{2}}}}\ln 4 $
Hence the correct option is (A) and (B).
So, the correct answer is “Option A and B”.

Note: Derivative is a small change in the value of a variable and the process of finding the derivative is known as differentiation. The most common example of it is change of displacement with respect to time which is known as velocity. The opposite of derivative is antiderivative. The derivative is found with respect to the independent variable. Every function has a different formula for finding the derivative.