Question

Differentiate: 1. \[y={{e}^{x}}\sec x-{{x}^{\dfrac{5}{3}}}\log x\]
2. \[y={{x}^{4}}+x\sqrt{x}\cos x-{{x}^{2}}{{e}^{x}}\]
3. \[y=\left( {{x}^{3}}-2 \right)\tan x-x\cos x+7{{x}^{7}}\]
4. \[y=\sin x\log x+{{e}^{x}}\cos x-{{e}^{x}}\sqrt{x}\]

Answer 1

Hint: In this problem, we have to differentiate the given differential equations. Here we can see that the equation is in addition and multiplication. So, we can use the formula \[\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}\] and \[\dfrac{d}{dx}\left( uv \right)=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\] to differentiate the given equations.

Complete step by step answer:
Here we have to differentiate each of the given equations.
We can see that the equations have only addition and multiplication in it, so we can use the formulas \[\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}\pm \dfrac{dv}{dx}\]…… (1), \[\dfrac{d}{dx}\left( uv \right)=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\]……. (2) for every problem.
 1. \[y={{e}^{x}}\sec x-{{x}^{\dfrac{5}{3}}}\log x\]
We can now differentiate the given equation
 We can now use the formula (2) in the first and the second term separately and add using the formula (1), we get
\[\Rightarrow \dfrac{dy}{dx}={{e}^{x}}\dfrac{d}{dx}\sec x+\dfrac{d}{dx}{{e}^{x}}\sec x-{{x}^{\dfrac{5}{3}}}\dfrac{d}{dx}\log x+\dfrac{d}{dx}{{x}^{\dfrac{5}{3}}}\log x\]
We can now use the differentiation formula for each term, we get
\[\Rightarrow \dfrac{dy}{dx}={{e}^{x}}\sec x\tan x+{{e}^{x}}\sec x-{{x}^{\dfrac{5}{3}}}\dfrac{1}{x}+\dfrac{5}{3}{{x}^{\dfrac{2}{3}}}\log x\]
We can now simplify the above term, we get
 \[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}={{e}^{x}}\sec x\left[ \tan x+1 \right]-{{x}^{\dfrac{2}{3}}}+\dfrac{5}{3}{{x}^{\dfrac{2}{3}}}\log x \\
\end{align}\]
Therefore, the answer is \[\dfrac{dy}{dx}={{e}^{x}}\sec x\left[ \tan x+1 \right]-{{x}^{\dfrac{2}{3}}}+\dfrac{5}{3}{{x}^{\dfrac{2}{3}}}\log x\].

2. \[y={{x}^{4}}+x\sqrt{x}\cos x-{{x}^{2}}{{e}^{x}}\]
We can now differentiate the given equation
 We can now use the formula (2) in the second and the third term separately and add and subtract using the formula (1), we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}{{x}^{4}}+\left( \dfrac{dx}{dx} \right)\sqrt{x}\cos x+x\cos x\left( \dfrac{d}{dx}\sqrt{x} \right)+x\sqrt{x}\left( \dfrac{d}{dx}\cos x \right)-\left[ {{x}^{2}}\dfrac{d}{dx}{{e}^{x}}+{{e}^{x}}\dfrac{d}{dx}{{x}^{2}} \right]\]
We can now use the differentiation formula for each term, we get
\[\Rightarrow \dfrac{dy}{dx}=4{{x}^{3}}+\sqrt{x}\cos x+x\cos x\dfrac{1}{2}\dfrac{1}{\sqrt{x}}+x\sqrt{x}\left( -\sin x \right)-\left[ {{x}^{2}}{{e}^{x}}+{{e}^{x}}2x \right]\]
We can now simplify the above term, we get
 \[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=4{{x}^{3}}+\sqrt{x}\cos x+\sqrt{x}\cos x\dfrac{1}{2}-x\sqrt{x}\sin x-{{x}^{2}}{{e}^{x}}-2x{{e}^{x}} \\
 & \Rightarrow \dfrac{dy}{dx}=4{{x}^{3}}-{{x}^{2}}{{e}^{x}}-2x{{e}^{x}}+\sqrt{x}\left( \cos x+\dfrac{1}{2}\cos x-x\sin x \right) \\
\end{align}\]
Therefore, the answer is \[\dfrac{dy}{dx}=4{{x}^{3}}-{{x}^{2}}{{e}^{x}}-2x{{e}^{x}}+\sqrt{x}\left( \cos x+\dfrac{1}{2}\cos x-x\sin x \right)\].
3. \[y=\left( {{x}^{3}}-2 \right)\tan x-x\cos x+7{{x}^{7}}\]
We can now differentiate the given equation
 We can now use the formula (2) in the first and the second term separately and add using the formula (1), we get
\[\Rightarrow \dfrac{dy}{dx}={{x}^{3}}\dfrac{d}{dx}\tan x+\tan x\dfrac{d}{dx}{{x}^{3}}-\dfrac{d}{dx}2\tan x-\left[ x\dfrac{d}{dx}\cos x+\cos x\dfrac{d}{dx}x \right]+\dfrac{d}{dx}7{{x}^{7}}\]
We can now use the differentiation formula for each term, we get
\[\Rightarrow \dfrac{dy}{dx}=3{{x}^{2}}\tan x+{{x}^{3}}{{\sec }^{2}}x-2{{\sec }^{2}}x-\left[ x\left( -\sin x \right)+\cos x \right]+49{{x}^{6}}\]
We can now simplify the above term, we get
 \[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=3{{x}^{2}}\tan x+{{x}^{3}}{{\sec }^{2}}x-2{{\sec }^{2}}x+x\sin x-\cos x+49{{x}^{6}} \\
 & \Rightarrow \dfrac{dy}{dx}=3{{x}^{2}}\tan x+{{\sec }^{2}}x\left( {{x}^{3}}-2 \right)+x\sin x-\cos x+49{{x}^{6}} \\
\end{align}\]
Therefore, the answer is \[\dfrac{dy}{dx}=3{{x}^{2}}\tan x+{{\sec }^{2}}x\left( {{x}^{3}}-2 \right)+x\sin x-\cos x+49{{x}^{6}}\].
4. \[y=\sin x\log x+{{e}^{x}}\cos x-{{e}^{x}}\sqrt{x}\]
We can now differentiate the given equation
\[\Rightarrow \dfrac{dy}{dx}=\sin x\dfrac{d}{dx}\log x+\log x\dfrac{d}{dx}\sin x+{{e}^{x}}\dfrac{d}{dx}\cos x+\cos x\dfrac{d}{dx}{{e}^{x}}-\left[ {{e}^{x}}\dfrac{d}{dx}\sqrt{x}+\sqrt{x}\dfrac{d}{dx}{{e}^{x}} \right]\]
We can now use the differentiation formula for each term, we get
\[\Rightarrow \dfrac{dy}{dx}=\sin x\dfrac{1}{x}+\log x\left( \cos x \right)+{{e}^{x}}\left( -\sin x \right)+\cos x{{e}^{x}}-\left[ {{e}^{x}}\dfrac{1}{2}\dfrac{1}{\sqrt{x}}+{{e}^{x}}\sqrt{x} \right]\]
We can now simplify the above term, we get
 \[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\sin x\left( \dfrac{1}{x} \right)+\log x\cos x+{{e}^{x}}\left( \cos x-\sin x \right)-{{e}^{x}}\left( \dfrac{1}{2\sqrt{x}}+\sqrt{x} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\sin x\left( \dfrac{1}{x} \right)+\log x\cos x+{{e}^{x}}\left( \cos x-\sin x \right)-{{e}^{x}}\left( \dfrac{1+2x}{2\sqrt{x}} \right) \\
\end{align}\]
Therefore, the answer is \[\dfrac{dy}{dx}=\sin x\left( \dfrac{1}{x} \right)+\log x\cos x+{{e}^{x}}\left( \cos x-\sin x \right)-{{e}^{x}}\left( \dfrac{1+2x}{2\sqrt{x}} \right)\].

Note: We should always remember the differentiating formulas such as \[\dfrac{d}{dx}\left( uv \right)=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\] and \[\dfrac{d}{dx}\left( u\pm v \right)=\dfrac{du}{dx}\pm \dfrac{dv}{dx}\]. We can use these formulas whenever necessary. We should also remember the formulas used for each of the terms and apply them correctly.