Original Address: https://zhaokaifeng.com/?p=1898
topic
Curve \ (\ sin (xy) + \ ln (yx) = x \) at the point \ ((0,1) \) tangent at the equation for the ____.
Resolve
That the need to use the knowledge derivation rules and tangent equation formula.
Need to use the derivation formula are:
\((\sin x)'=\cos x;\)
\((\ln x)'=\frac{1}{x};\)
\((ab)'=a'b+ab';\)
\(f'(x)=f'[\phi(x)]\cdot\phi'(x).\)
Derivation process also important to note the following points:
For \ (X \) derivative, include \ (X \) and other constants to be calculated according to the equation derivation, and in addition to \ (X \) other than the variable derivation of only plus symbols (e.g.: ') can not be calculated derivation;
After the derivation of the same variable on both sides of the equation, the equation still holds. Because the first derivative of equation, derivation rules are the same, then the derivative still identities on both sides and so on.
Equation tangent calculated as follows:
\(y-f(x_{0})=f'(x_{0})(x-x_{0}).\)
Ideas to answer the following:
Since the calculated tangential equation contains the derivative \ (F '(X) \) , therefore, first need to compute the derivative. Sides while the original formula \ (X \) derivation may be generated derivative \ (Y '\) :
\([\sin(xy)+\ln(y-x)]'=(x)'\Rightarrow\cos(xy)(x'y+xy')+\frac{1}{y-x}(y-x)'=1\Rightarrow \cos(xy)(y+xy')+\frac{1}{y-x}(y'-1)=1\)
Requirement is a graph at the point \ ((0,1) \) tangent at the equation, therefore, we \ (x = 0; y = 1 \) to the above into the equation to give:
\(1\cdot1+1\cdot(y'-1)=1\Rightarrow 1+y'-1=1\Rightarrow y'=1.\)
which is:
\ (Y '(0) = 1. \)
The above results into equation derivation tangent formula obtained:
\(y-1=1\cdot(x-0)\Rightarrow y=x+1.\)
To sum up, the answer to this question is obtained: \ (Y = X +. 1 \)
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