Chapter 3: Problem 15
Find a point \(x\) where \(f^{\prime}(x)=6\). $$ f(x)=x^{2}+5 $$
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Chapter 3: Problem 15
Find a point \(x\) where \(f^{\prime}(x)=6\). $$ f(x)=x^{2}+5 $$
These are the key concepts you need to understand to accurately answer the question.
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Differentiate the given expression with respect to \(x\). $$ x \arcsin (x) $$
\- A real-valued function \(f\) of a real variable \(x\) is said to be algebraic if there is a polynomial \(p(u, v)\) with integer coefficients such that \(p(x, f(x))=0\) for all \(x .\) For example, \(f(x)=\) \(2 x+\sqrt{x^{2}+1}\) is algebraic because $$ \begin{aligned} p(x, f(x))=& 3 x^{2}+\left(2 x+\sqrt{x^{2}+1}\right)^{2} \\ &-4 x\left(2 x+\sqrt{x^{2}+1}\right)-1 \\ \equiv & 0 \end{aligned} $$ for \(p(u, v)=3 u^{2}+v^{2}-4 u v-1 .\) A function that is not algebraic is said to be transcendental. Find a polynomial that shows that the given expression is algebraic. $$ \sqrt[3]{x+\sqrt{x}} $$
\- A real-valued function \(f\) of a real variable \(x\) is said to be algebraic if there is a polynomial \(p(u, v)\) with integer coefficients such that \(p(x, f(x))=0\) for all \(x .\) For example, \(f(x)=\) \(2 x+\sqrt{x^{2}+1}\) is algebraic because $$ \begin{aligned} p(x, f(x))=& 3 x^{2}+\left(2 x+\sqrt{x^{2}+1}\right)^{2} \\ &-4 x\left(2 x+\sqrt{x^{2}+1}\right)-1 \\ \equiv & 0 \end{aligned} $$ for \(p(u, v)=3 u^{2}+v^{2}-4 u v-1 .\) A function that is not algebraic is said to be transcendental. Find a polynomial that shows that the given expression is algebraic. $$ \sqrt{x}+\sqrt{3} $$
Differentiate the given expression with respect to \(x\). $$ \tanh ^{-1}(\cos (x)) $$
Suppose that \(f, g,\) and \(h\) are differentiable functions and that \(f=g \cdot h\). Suppose that \(g(4)=6, d g(4)=7, d x, h(4)=\) \(5,\) and \(d h(4)=-2 d x .\) What is \(d f(4) ?\)
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