Practice implicit differentiation problems & examples, Learn equation derivatives at particular values of ?x? and ?y? by implicit differentiation & higher order derivatives of simple function given in the implicit form, orthogonal curves.
Functions are many times defined in a way that \(y\) is expressed clearly in terms of \(x\) , we say \(y\) is an explicit function of \(x\) .
\(y =f(x) \to \) explicit function
e.g \(x^2+y^3=3xy\to\) implicit relation
We say \(y\) is an implicit function of \(x\) in this case.
and solve the resulting equation for \(y'\) .
So \(\dfrac{d}{dx} (f(x))^n = n(f(x))^{n-1} f'(x) \) by Chain Rule.
A \(\dfrac{46}{1-y^2}\)
B \(\dfrac{2}{7(1-y^2)}\)
C \(2y+3\)
D \(\dfrac{4}{3(1-y^2)}\)
A \(\dfrac{sin\,x^3 - cos\,x^3}{x^2-y^2}\)
B \(\dfrac{sin\,y^3 + 3y\,cos\,x^3}{2\,y^2+x^3}\)
C \(\dfrac{sin\,y^3 - 3\,x^2y\,cos\,x^3}{sin\,x^3 - 3y^2\,x\,cos\,y^3}\)
D \(\dfrac{cos\,y^3 - 3\,y^2x\,cos\,x^3}{sin\,x^3 +y^3}\)
A \(\dfrac{-2}{3}\)
B \(-4\)
C \(6\)
D \(8\)
If higher order derivatives of implicit function are desired , we do the following steps:
(1) Differentiate both sides with respect to \(x \), treating \(y\) as a function of \(x \).
(2) Again differentiate both sides with respect to \(x \) , taking all terms of \(y'\) obtained in (1) as function of \(x \) .
(3) Obtain an expression for \(y'' \) , if it contains \(y'\) term. Use result of step (1) to express \(y'' \) in terms of \(x \) and \(y\) .
A \(\dfrac{-4\,x^2-6\,y^2}{9\,y^3}\)
B \(\dfrac{4\,x^2+6\,y^2}{\,x^3}\)
C \(\dfrac{7\,x^2+\,y^2}{\,y^3}\)
D \(\dfrac{2\,x^2-6\,y^2}{5\,y^3}\)
A \(4x+7y-1=0\)
B \(4\,x+3\,y-2=0\)
C \(2\,x+3\,y-1=0\)
D \(2\,x+\,y-3=0\)
\(x^2y''+2\,x\,y' + 2\,x\,y' +2\,y =0\) \(\Rightarrow x^2\,y'' +4\,xy'+2\,y=0\)
A \(\dfrac{3y}{4x^2}\)
B \(\dfrac{-(x^3+4x)}{4\,y^2}\)
C \(\dfrac{-(x+x\,y^2)}{\,x}\)
D \(\dfrac{-(y+y^3)}{5\,x}\)
\( y= f(x) \) and \( y= g(x) \) are orthogonal , tangent are \(\bot\) at the intersection point P.
slope of \(\ell_2 = tan\,\theta_2 = m_2\)
Now \(\theta_2 = 90° +\theta_1\)
\(\Rightarrow tan\,\theta_2 = tan(90°+\theta_1)\,\)
\(\Rightarrow tan\,\theta_2 = -cot\,\theta_1\)
\(\Rightarrow tan\,\theta_2 = \dfrac{-1}{tan\,\theta_1} \)
\(\Rightarrow 1+tan\,\theta_2 \,tan\,\theta_1=0\,\)
\(\Rightarrow 1+m_1m_2=0\)
\(\Rightarrow\, m_1m_2 = -1\)
(for \(\bot\) lines \(m_1\,m_2=-1\))
A \(x^2+2y^2 =1\)
B \(x^2=6y\)
C \(y^2=6\,x\)
D \(3\,y^2+x^2=1\)
\(3(f(x))^2 \,f'(x) + f'(x) =10 \to\) put \(x=1\) .
\(\Rightarrow 3× (f(1))^2 × f'(1)+f' (1) = 10\)
\(\Rightarrow 3× 4× f' (1) + f' (1) = 10\)
\(\Rightarrow f'(1) = \dfrac{10}{13}\)
A \(\dfrac{-2}{5}\)
B \(\dfrac{-3}{7}\)
C \(81\)
D \(\dfrac{4}{7}\)