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Sagot :
Hello there. To solve this question, we'll have to remember some properties about derivatives.
Given the following function:
[tex]f(x)=\dfrac{5\tan(x)}{x}[/tex]We want to take its derivative.
For this, we'll have to apply some rules:
The derivative of a quotient of two continuously differentiable functions is given by the quotient rule:
[tex]\left(\dfrac{f(x)}{g(x)}\right)'=\dfrac{f'(x)\cdot\,g(x)-f(x)\cdot\,g'(x)}{(g(x))^2}[/tex]The derivative of the tangent function can be found using the above rule and by knowing the derivatives of sine and cosine. In this case, we obtain the following:
[tex]\begin{gathered} (\tan(x))^{\prime}=\left(\dfrac{\sin(x)}{\cos(x)}\right)^{\prime}=\dfrac{(\sin(x))'\cdot\cos(x)-\sin(x)\cdot(\cos(x))'}{(\cos(x))^2} \\ \\ \Rightarrow\dfrac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=\dfrac{1}{\cos^2(x)}=\sec^2(x) \end{gathered}[/tex]The derivative of a power can be found using the limit definition of a derivative, in this case we obtain:
[tex](x^n)^{\prime}=n\cdot\,x^{n-1},\text{ }n\in\mathbb{R}[/tex]With these rules, we can already solve the question.
First, the derivative is a linear operator, which means that
[tex](\alpha\cdot\,f(x))^{\prime}=\alpha\cdot\,f^{\prime}(x),\text{ }\alpha\in\mathbb{R}[/tex]Hence we get that
[tex]f^{\prime}(x)=\left(\dfrac{5\tan(x)}{x}\right)^{\prime}=5\cdot\left(\dfrac{\tan(x)}{x}\right)^{\prime}[/tex]Now, apply the quotient rule
[tex]f^{\prime}(x)=5\cdot\dfrac{(\tan(x))'\cdot\,x-\tan(x)\cdot(x)'}{x^2}[/tex]Take the derivative of the tangent and apply the power rule
[tex]f^{\prime}(x)=5\cdot\dfrac{\sec^2(x)\cdot\,x-\tan(x)\cdot1}{x^2}[/tex]Therefore we get that
[tex]f^{\prime}(x)=5\cdot\dfrac{x\sec^2(x)-\tan(x)}{x^2}[/tex]Is the answer to this question.
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